Interactive software for chemistry students and chemists

How has the learning software developed since the introduction of the PC in the early 80's. Has the potential of the computer as a relatively new educational tool been fully understood? CHEMIX was developed as a serious attempt to improve the quality of learning chemistry by the use of computers.
CHEMIX is an interactive educational software tool developed by a chemist (and teacher) for students (high schools, colleges) teachers and chemists. As a stand-alone product it covers a wide range of topics in the area of chemistry.
By the property of interactivity CHEMIX will instantaneously respond to inputs. As an educational tool it will increase the frequency of the repeated process - trial and error - the most common way of learning.
A popular place for using CHEMIX is in the laboratory, where it functions as a standard calculation tool. This is a place where time consuming and often repeated manual calculations with advantage can be replaced by a tool in which performs efficient and secure calculations.
In the classroom CHEMIX is useful for students who after learning to do problems (in which also are included) by hand, can use CHEMIX to verify their results. It can even be used for correcting erroneous calculations and answers in which often are found in the literature.
The problems/lessons supplied with this software, are intended and leveled for the first-year university level, many of the lessons are also suitable for high school courses.
The thorough understanding of the physical and chemical principles involved in a problem is essential in order to apply intelligently the mathematics used in the solution of the problems. Therefore, it is provided to have some basic chemical and physical knowledge before fully taking advantage of CHEMIX learning and censorial capability.

103 elements and their stable isotopes activated by push buttons
22 physical properties shown simultaneously in edit fields activated by radio buttons.
Graphics

1) History
2) 22 physical properties (melting point, boiling point, electronegativity .....)
3) Stable isotopes
4) Natural abundance
5) Spin
6) Atomic mass
7) Magnetic moment
8) Quadrupole moment
9) Resonance frequency
10) Relative receptivity
11) Magnetogyric ratio
12) Physical properties of more than 2500 unstable isotopes:
Atomic mass, Half life, Decay modes, Decay energy, Particle energy,
Particle intensities, Spin and Magnetic moment.
13) More than 600 decays (decay trees)
14) Magnetic susceptibility

1) 103 elements
2) all stable isotopes
3) deuterium(e.g.D2O)
4) ions
5) braces (multilevel)
6) arg. calculations (mol-mass conv.)
7) calculations argumented by an element
8) crystal water
9) error guidance

Tab. Molecular Calculator (signs and symbols)

Sign/symbol	Example	Interpretation
m	2mH2O	mol
g	2gH2O	gram
E or e	2E-3mH2O or 2e-3mH2O	exponent
e-	e-	electron symbol (normally used by "balance")
,	2mH2,H2O	argument separator
.	2.3gH2O	decimal separator
*	CaCl2*5H2O	crystal water separator
D	D2O or H[2]2O	deuterium (symb. reserved for this isotope)
[ ]	H[2] or [Fe(CN)6]+2	number of nucleons square brackets (isotopes and complex)
( )	Ca(NO3)2	braces
+	Ca+2 or Ca++	pos. charged ion
-	SO4-2 or SO4--	neg. charged ion
0-9	10gH2O	numbers
H-Lr	HHeLiBeBCNO	Elements: Hydrogen --> Lawrencium

Tab. Explore Molecular Calculator

Normal	Chemix	Interpretation
H₂SO₄ and NO₃^-	H2SO4 and NO3-	Index
H₂SO₄	H2SO4	Formula mass(weight)
C[13]₅H[2]₁₂	C[13]5H[2]12	Enriched
2 moles of H2SO4	2mH2SO4	Arg. formula (mol)
2 grams of H2SO4	2gH2SO4	Arg. formula (mass)
2 moles of sulfur in H2SO4	2mS,H2SO4	Formula arg. by element and mol
2 grams of sulfur in H2SO4	2gS,H2SO4	Formula arg. by element and mass
2 moles of H₂ in H₂SO₄	2mH2,H2SO4	Formula arg. by molecuel and mol
2 grams of H₂ in H₂SO₄	2gH2,H2SO4	Formula arg. by molecuel and mass
2 moles of deuterium in H[2]2SO4	2mH[2],H[2]2SO4	Enriched formula arg. by isotope and mol
2 grams of deuterium in H[2]2SO4	2gH[2],H[2]2SO4	Enriched formula ar by isotope and mass
CaCl₂*5H2O	CaCl2*5H2O	Crystal water comp.
2 moles of CaCl₂*5H2O	2mCaCl2*5H2O	2 moles of cryst.water comp.
[Fe(CN)₆]⁺²	[Fe(CN)6]+2	Complex brackets
0.025 moles of C₄H₁₀	2.25E-2mC4H10	Use of exponent in arg.
SO₄^-2 or SO₄^2- or SO₄ ^--	SO4-2	Ions

Tab. Examples from program

Examples	Chemix	Interpretation
1)	H2SO4	Formula mass sulfuric acid
2)	H[2]2SO4 or D2SO4	Formula mass sulfuric acid (enriched by deuterium)
3)	2gH2SO4	2 grams of sulfuric acid
4)	2mH2SO4	2 moles of sulfuric acid
5)	2gS,H2SO4	2 grams of sulfur in sulfuric acid
6)	2mS,H2SO4	2 moles of sulfur in sulfuric acid

CHEMIX School concentration calculator offer an easy and intuitive way of calculate and convert between different unit concentrations. It also allow dilution calulations using a stock standard solution. It can convert/calculate the weight(g) of a substance to mol and vice versa. By using the substance density database it also has the ability to convert/calculate between volume(mL) and the mass(g) of the solution.

By first selecting a substance and by inserting two legal values in (mass or mol) and (V or m_s) "Amount" fields
the concentration calculator vil calculate 5 different types of concentrations + the density of the solution (Concentration frame).

CHEMIX School concentration and dilution calculator allow efficient dilution calulations. A minimal amount of steps is needed for this kind of calculations.
It is possible to use molar, molal, w/w%, w/v% or mole fraction (x) stock concentrations .
Useful formulas: C₁ V₁ = C₂ V₂ M = n_solute/V_solution m = n_solute / m_solvent

Knowing that the density of a 17.4M CH₃COOH solution = 1.05148 g/mL
a) How many mL of a 17.4M a acetic acid stock solution is needed for making 3500 mL of a 2M acetic acid solution.
b) How many grams of a 17.4M a acetic acid stock solution is needed for making 3500 mL of a 2M acetic acid solution.

We know that: M = n/V_solution C₁ V₁ = C₂ V₂ (dilution formula)
Solution a)

17.4M * C₂ = 3.5L * 2M

V₂ = (3.5L * 2M)/17.4M = 0.4023 L = 402.3mL

Solution b)

Since we already know (from solution a) ) the volume of a 17.4M solution needed and the density we kan calculate the mass :

We need measure out 402.3mL * 1.05148 g/mL= 423.01g of 17.4M acetic acid

The molality (m)of a solute is the number of moles of that solute divided by the weight of the solvent in kilograms. For water solutions, 1 kg of water has a volume close to that of 1 liter, so molality and molarity are similar in dilute aqueous solutions.
Molality (m) is typically used in thermodynamic calculations where a temperature independent unit of concentration is needed. Since density is a temperature dependent property a solution’s volume, and thus it’s eg. molar concentration will change as a function of its temperature. By using the solvent’s mass in place of its volume (molal vs molar), the resulting concentration becomes independent of temperature.

Solvent : the liquid in which a solute dissolves
Solute : the substance that dissolves in a liquid to form a solution
Solution : is the mixture formed when a solute has dissolved in a solvent

moles HNO₃ = 70.4 g / 63.0129 g/mol = 1.11723 mol
kg of water = 0.0296 kg
molality = 1.11723 mol / 0.0296 kg = 37.744 m

Balance, step by step Problem 1: Balance the equation for combustion of penthane

Step 1. Insert 1 in front of the most complicated looking compound.
1C₅H₁₂ + O₂ --> CO₂ + H₂O

Step 2. To balance C's, 5 must be inserted in front of CO₂.
1C₅H₁₂ + O₂ --> 5CO₂ + H₂O

Step 3. To balance H's, 6 must be inserted in front of H₂O .
1C₅H₁₂ + O₂ --> 5CO₂ + 6H₂O

Step 4. To balance O's, 8 must be inserted in front of O₂ .
1C₅H₁₂ + 8O₂ = 5CO₂ + 6H₂O

Erase 1 from the equation:

C₅H₁₂ + 8O₂ = 5CO₂ + 6H₂O

The equation is now balanced!
Note: If any fractions should occur, simply multiply the equation by a suitable number. This will be demonstrated in the next problem.

Problem 2: Balance the equation for combustion of ethane

Step 1. Insert 1 in front of the most complicated looking compound.
1C₂H₆ + O₂ --> CO₂ + H₂O

Step 2. To balance C's, 2 must be inserted in front of CO₂.
1C₂H₆ + O₂ --> 2CO₂ + H₂O

Step 3. To balance H's, 3 must be inserted in front of H₂O .
1C₂H₆ + O₂ --> 2CO₂ + 3H₂O

Step 4. To balance O's, 7/2 must be inserted in front of O₂ .
1C₂H₆ + 7/2O₂ = 2CO₂ + 3H₂O

Step 5. Eliminate the fraction 7/2. This can be done by multiplying the equation by 2.
2C₂H₆ + 7O₂ = 4CO₂ + 6H₂O

The equation is now balanced!

Tab. Balance (signs and symbols)

Sign/symbol	Example	Interpretation
+	H2 + O2 ....	Separates compounds in equation
>	H2 + O2 > H2O	Separates left and right side (normally an arrow)
e-	Cu + H2O > Cu2O + H+ + e-	Electron

Examples from program - Balance:

1) FeS2+O2>Fe3O4+SO2
2) C3H8+O2>CO2+H2O and calculate the amount of CO2 and H2O formed if 2 grams of C3H8 (propane) reacts with O2.
3) C3H8+O2>CO2+H2O and calculate the amount of CO2 and H2O formed if 2 moles of C3H8 (propane) reacts with O2.
4) C3H8+O2>CO2+H2O and calculate the amount of CO2 and H2O formed if 5 grams of C3H8 (propane) reacts with 3 grams of O2.
5) C3H8+O2>CO2+H2O and calculate the amount of CO2 and H2O formed if C3H8 (propane) containing 2 grams of H reacts with 3 grams of O2.
6) As2S3+NaNO3+Na2CO3>Na3AsO4+Na2SO4+NaNO2+CO2 and calculate the amount of As2S3 used forming CO2 containing 10 grams of carbon.

Every chemical change and every physical change involves an energy change. These energy changes that accompany the transformation of matter help us understand better the nature of chemical and physical changes.

First law of thermodynamics : Energy can be neither created nor destroyed.

Chemical thermodynamics is the study of the energy effects accompanying chemical and physical changes. Three of these thermodynamic quantities are: enthalpy, H, entropy, S, and Gibbs free energy, G.

The enthalpy of formation of a compound is a change in energy that occurs when it is formed from its element.

CH₄ + 2O₂ --> CO₂ + 2H₂O ,

H = -890.2 kJ/mol
4Fe(s) + 3O₂(g) --> 2Fe₂O₃(s) ,

H = -1647 kJ/mol
N₂O₄(g) --> 2NO₂(g) ,

H = 57.24 kJ/mol

The enthalpies of formation may be either positive or negative. A positive enthalpy of formation indicates that energy has to be provided in order for the reaction to proceed. A negative enthalpy of formation indicates that energy is evolved.
What the enthalpy of formation tells us is the change of enthalpy that would occur if a reaction pathway could be found that takes one of the elements to the compound of interest.

Hess's law: The overall reaction enthalpy is the sum of reaction enthalpies of each step into which the reaction may formally be divided.

The entropy, S, of a system is a measure of its randomness or, its disorder.

Second law of thermodynamics: The entropy of the universe increases in the course of every natural change.

Consider the oxidation of iron (25^oC).:
4Fe(s) + 3O₂(g) --> 2Fe₂O₃(s)

S = 2(87.37J/(K mol)) - 4(27.29J/(K mol)) - 3(205.2J/(K mol)) = -550.12J/(K mol) (

H = -1647kJ/mol)
The negative enthalpy in this reaction (exothermic reaction), causes the

S(surroundings) to be positive:

S(surroundings) = -

H/T

S(surroundings) = -(-1647 kJ/mol)/(298.15 K) = +5524 J/(K mol)
This increase in entropy arises because the reaction releases energy into the surroundings.

In contrast to oxidation of iron, the reaction: N₂O₄(g) --> 2NO₂(g)
is an endothermic reaction (

H = +57.2 kJ/mol):

S = 2(240.1J/(K mol)) - 304.38J/(K mol) = 175.82J/(K mol)

S(surroundings) = -(57.24 kJ/mol)/(298.15 K) = -192 J/(K mol)
All endothermic reactions decrease the entropy of the surroundings.

The sum of the total entropy changes

S(total) = -

H/T +

S(surroundings) is a measure of the total change of energy in the universe.
For every 4 mol of Fe converted to 2 mol Fe₂O₃ there is an overall increase in the entropy of the universe:

S(total) = +5524J/(K mol) + (-550.12J/(K mol)) = +4973.88 J/(K mol)
This large positive quantity indicates that the reaction is spontaneous and a thermodynamic reason why steel corrodes.

The dissociation of N₂O₄ indicates that the total entropy change:

S(total) = -192J/(K mol) + (175.82J/(K mol)) = -16.18 J/(K mol)
Because this is a small, negative value, we can conclude that this reaction does not go to completion.

When a reversible chemical change occurs, the difference between

H and T

S represents the amount of available energy released or absorbed as a result of the total change.

The new quantity -T

S(total) is now given a new symbol and a new name. It is called the Gibbs function of reaction or Gibbs free energy and denoted

The last equation therefore becomes:

G =

H - T

S

This energy represents the driving force of the reaction. When

G is negative the reaction will be a spontaneous one; when

G is positive the reverse of the reaction as written will be the spontaneous one; when

G is zero equilibrium exists. The magnitude of

G is a measure of the extent to which the reaction will go to completion. Thus, a knowledge of

G values enables us to predict the course of a reaction. For example, we can predict that the reaction: Cl2 (g) + 2I-(aq) --> 2Cl-(aq) + I2(g)

G = -38 kcal will be a spontaneous one since

G is negative. Had

G been positive, the spontaneous reaction would have been the reverse of the reaction as written.

Tab. Thermochemistry (state symbols)

State symbols	Example	Interpretation
(g)	H2(g)	Gas
(l)	H2O(l)	Liquid
(s)	Cu(s)	Solid
(aq)	Cu+2(aq)	Aqua

Ammonium nitrate ,NH4NO3, decomposes explosively:
NH4NO3(s) > N2O(g) + H2O(g)
a) Calculate

H for the balanced reaction.
b) If 36 grams of H2O are formed from the reaction, how much heat was released?

Solution:
a) Insert and calculate: NH4NO3(s) > N2O(g) + H2O(g)
b) Arg. H2O by 36g, insert and calculate: NH4NO3(s) > N2O(g) + 36gH2O(g)

Here's an example of an equation using all the state symbols (g),(l),(s) and (aq)
CaCO3(s) + H2SO4(aq) > Ca+2(aq) + CO2(g) + SO4-2(aq) + H2O(l)

Some compounds dissolve in water as molecules while others, called electrolytes, dissociate and dissolve as charged species called ions. Compounds which exist as solid ionic crystals are mostly highly soluble in water. Ionic compounds dissolve to the point where the solution is saturated and no more solid can dissolve. The concentration of the saturated solution is termed the solubility of the substance. In some cases the solubility may be very high and a large amount of the solid may dissolve before the solution is saturated. The difference in the ability to dissolve in water are large. Some highly soluble salts dissolve very easily (1/2 kg in 1 kg water). Other salts doesn't seem to dissolve at all (1*10^-15 g in 1 kg water), even if the temperature is the same. The description "highly soluble" generally means soluble to at least the extent of forming 0.1 to 1.0 molar aqueous solutions. Salts which are less soluble in water than this at room temperature are called slightly soluble salts.

1) Salts of alkali metals (group I) are highly soluble. Exception is KClO₄ (moderately soluble)
2) Nitrates and ammonium salts.
3) Metal halides are generally highly soluble. Exceptions are those of : Pb⁺², Ag⁺ og Hg₂⁺².
4) Most sulfate salts. Exceptions are those of: Ca⁺², Ba⁺², Sr⁺², Pb⁺² and Hg₂⁺².

1) Salts of carbonates, phosphates, hydroxides and sulfides are usually insoluble. Exceptions are alkali metals (group I) and following moderately soluble salts: Ca⁺², Ba⁺², Sr⁺².
2) Metal sulfides are generally insoluble. Solubility Product Constant
Ionic compounds dissolve to the point where the solution is saturated and no more solid can dissolve. The concentration of the saturated solution is termed the solubility of the substance. In such a saturated solution, equilibrium is established. Ions will form solid in the same extent as solid dissociate and form ions. The solubility product constant or solubility product K_sp is a temperature-dependent constant referring to this stage.
If M_xA_y dissociate into (cations) M^+m and (anions) A^-a The expression for solubility product will be:

K_sp=[M^+m]^x [A^-a]^y
The Common Ion Effect
The concentrations of ions in solution are affected by all equilibria and all species present in the solution. The simplest and most significant such effect is called the common ion effect. The common ion effect is observed whenever an ion in solution is common to two different salts which serve as its sources. Addition of the second salt adds the common ion, which is a product of the dissolution of the first. The effect of adding the product ion will be to decrease the solubility of the first salt.
If a salt M1_xA1_y (e.g. BaF2 : x=1, y=2) is added into a solution already containing a common ion e.g. M2^+m (Ba⁺²), the expression for the solubility product will be:

K_sp=[M1^+m+M2^+m]^x [A1^-a]^y

Ex.1:How many moles of AgCl will dissolve in 0.5 kg of water? K_sp(AgCl)=1.77E-10.
Solution:
1) Insert equation of dissociation: AgCl = Ag+ + Cl-
2) Insert 1.77E-10 in K_sp-field
3) Insert mass (solvent): 0.5 kg
4) Calculate

Ex.2: 1.07722E-17 moles of Ag₂S will dissolve in 1 kg of water. Calculate K_sp(Ag₂S)
Solution:
1) Insert equation of dissociation: Ag₂S = 2Ag+ + S-2
2) Insert 1.07722E-17 in Ag₂S-field
3) Insert mass (solvent): 1 kg
4) Calculate

Ex.3: How many grams and moles of BaSO₄ (K_sp=1.8E-10) will dissolve in 1000 kg of water?
Solution:
1) Insert equation of dissociation: BaSO₄ = Ba+2 + SO4-2
2) Insert 1.8E-10 in K_sp-field
3) Insert mass (solvent): 1000 kg
4) Calculate

Ex.4: How many grams of BaSO₄ (K_sp=1.8E-10) will dissolve in a 1000 kg 0.1M Na₂SO₄-solution ?
Solution:
1) Insert equation of dissociation: BaSO₄ = Ba+2 + SO4-2
2) Insert 1.8E-10 in K_sp-field
3) Insert Na₂SO₄-conc. (0.1) in SO₄^-2-common ion effect field.
4) Insert mass (solvent): 1000 kg
5) Calculate

Ex.5: How many grams of MgF₂ (K_sp=7.42E-11) will dissolve in 0.5 kg of water?.
Solution:
1) Insert equation of dissociation: MgF₂ = Mg+2 + 2F-
2) Insert 7.42E-11 in K_sp-field
3) Insert mass (solvent): 0.5 kg
4) Calculate

Ex.6: How many grams of MgF₂ (K_sp=7.42E-11) will dissolve in a 0.5 kg 0.1M NaF-solution?.
Solution:
1) Insert equation of dissociation: MgF₂ = Mg+2 + 2F-
2) Insert 7.42E-11 K_sp-field
3) Insert (0.1) in F^--conc. common ion field.
4) Insert mass (solvent): 0.5 kg
5) Calculate

Solutions 1-6 Ex.1: 6.652E-6 mol
Ex.2: 5E-51
Ex.3: 3.13 g , 0.013 mol
Ex.4: 0.00042 g
Ex.5: 0.00825 g
Ex.6: 2.31E-7 g

Weak acids are compounds that partially dissociate to produce an equilibrium concentration of H⁺. An example equilibrium is acetic acid in water :

Weak bases partially react with water to produce an equilibrium concentration of OH^-. An example equilibrium is ammonia in water:

The [CH₃COOH] started at 0.2 M and went down as CH₃COOH molecules dissociated.

[H₃O⁺] = [CH₃COO^-] = x

The [CH₃COOH] started at 0.2 M and went down as CH₃COOH molecules dissociated. In this approximation we simplify by ignoring that molecules dissociate from the original conc. substituting (0.2-x) by 0.2, since x is rather small.

What is the pH of a 0.200 M solution of acetic acid (K_a = 1.75E-5)

1) Insert equation of dissociation: CH₃COOH + H₂O > CH₃COO^- + H₃O⁺
2) Insert 1.75E-5 in K-field
3) Insert 0.2 in CH₃COOH (Before dissoc.)-field
4) Calculate

Ex. 1: Calculate pH and H₃O⁺-conc. in a 0.1M solution of acetic acid(K_a=1.75E-5).
Solution:
1) Insert equation of dissociation: CH₃COOH + H₂O > CH₃COO^- + H₃O⁺
2) Insert 1.75E-5 in K-field
3) Insert 0.1 in CH3COOH (Before dissoc.)-field
4) Calculate

Ex. 2: Decide the amount of undissociated acetic acid in a [H₃O⁺] = 0.002M solution (K_a=1.75E-5)
. Solution:
1) Insert equation of dissociation: CH₃COOH + H₂O > CH₃COO^- + H₃O⁺
2) Insert 1.75E-5 in K-field
3) Insert 0.002 in H₃O⁺/OH^--field
4) Calculate

Ex. 3:Decide conc. of dissoc. and undissoc. acetic acid in a pH=4 solution (K_a=1.75E-5)
Solution:
1) Insert equation of dissociation: CH₃COOH + H₂O > CH₃COO^- + H₃O⁺
2) Insert 1.75E-5 in K-field
3) Insert 4 i pH-field
4) Calculate

Ex. 4:Find pH in a solution containing 0.1M CH₃COONa and 0.1M acetic acid(K_a=1.75E-5)
Solution:
1) Insert equation of dissociation: CH₃COOH + H₂O > CH₃COO^- + H₃O⁺
2) Insert 1.75E-5 in K-field
3) Insert 0.1 in CH₃COOH (Before dissoc.)-field
4) Insert 0.1 in CH₃COO^- common ion field
5) Calculate.

Ex. 5:a)How many moles of NaAc (CH₃COONa) must be added in a 1 dm³ 0.001M CH₃COOH-solution making a pH=7.0-buffer (K_a=1.75E-5).
Solution:
1) Insert equation of dissociation: CH₃COOH + H₂O > CH₃COO^- + H₃O⁺
2) Insert 1.75E-5 in K-field
3) Insert 0.001 in CH₃COOH (Before dissoc.)-field
4) Insert 7.0 pH-field
5) Calculate

Ex. 6:How many moles of NaCN must be added in a 1 dm³ 0.2M HCN before [H₃O⁺] = 1.0E-6 ? (K_a=5.85E-10)
Solution:
1) Insert equation of dissociation: HCN + H₂O > CN^- + H₃O⁺
2) Insert 5.85E-10 in K-field
3) Insert 1E-6 in H₃O⁺-field
4) Insert 0.2 i HCN (before dissoc.) field.
5) Calculate

Solutions Ex.1: pH=2.88, H₃O⁺-conc.=0.001314M
Ex.2: 0.2286M
Ex.3: Dissoc=0.0001M, un-dissoc=0.000571M
Ex.4: pH=4.757
Ex.5: 0.1745 mol
Ex.6: 0.000116 mol

Electrochemistry is the study of interchange of chemical and electrical energy. Oxidation/Reduction (redox) involves the exchange of electrons from one chemical species to another. Normally, this is done when the two chemicals contact each other (bump into each other). Separating the chemical species such that the electrons transfer via an external circuit, we can measure the electrochemical effects.

Redox, reduction-oxidation reactions are reactions that involves transfer of electrons. In the following example we will see that a redox reaction involves both reduction and oxidation:
Example:
2Br^- --> Br₂ + 2e^- : Oxidation
Cl₂ + 2e^- --> 2Cl^- : Reduction
-------------------------------------------
2Br^- + Cl₂ = Br₂ + 2Cl^- : Redox reaction

Reduction: the donation of electrons to a species.
Oxidation: the removal of electrons from a species.

A substance which causes another to get oxidized is called an oxidizing agent (or oxidant) and will itself get reduced.
A substance which causes another to get reduced is called a reducing agent (or reductant) and will itself get oxidized.

Mnemonic: Oxidation Is Loss, Reduction Is Gain -- (OIL RIG)

An oxidation state change indicates how many electrons transferred per species.

1) Oxygen in compounds is assigned an oxidation state of -2. (Exception: peroxides, e.g. H₂O₂)
2) Hydrogen in compounds is assigned an oxidation state of +1. (Exception: hydrides, e.g. NaH, KH ...)
3) Free elements such as e.g. O₂ and Na are assigned an oxidation state of zero.
4) The sum of the oxidation states of all the atoms in a species must be equal to the net charge on the species.
5) The alkali metals (Li, Na, K, Rb, and Cs) in compounds are always assigned an oxidation state of +1.
6) The alkaline earth metals(Be, Mg, Ca, Sr, Ba, and Ra) and also Zn and Cd in compounds are always assigned an oxidation state of +2.

7) In an acid solution, use H⁺ and H₂O to balance charges and other atoms. In a basic solution, use OH^- and H₂O to balance charges and other atoms.

By using these guidelines we can figure out oxidation states for all elements involved in a redox reaction.

Examples from program - problems and solutions
Example 1 Balance: Fe⁺² + MnO₄^- + H⁺ --> Fe⁺³ + Mn⁺² + H₂O
Solution:
The oxidation state of oxygen is -2 (Guideline 1). By knowing this we can decide oxidation state of manganese in MnO₄^- by the use of guideline 4 (net charge).
The oxidation state of manganese must be +7 because : 4*(-2)+7 = -1 (net charge)

Iron: Fe⁺² - e^- --> Fe⁺³ Multiply by 5
Manganese: Mn⁺⁷ + 5e^- --> Mn⁺² Multiply by 1
------------------------------------------------
Overall: 5Fe⁺² + Mn⁺⁷ --> 5Fe⁺³ + Mn⁺²

The other species in the equation can now be balanced by inspection.

5Fe⁺² + MnO₄^- + H⁺ --> 5Fe⁺³ + Mn⁺² + H₂O

5Fe⁺² + MnO₄^- + 8H⁺ --> 5Fe⁺³ + Mn⁺² + 4H₂O

CHEMIX solution:
Insert: Fe⁺² + MnO₄^- + H⁺ > Fe⁺³ + Mn⁺² + H₂O
in one of the half-reaction fields and calculate.
NOTE: In this case only one of the two fields should contain an equation.
Example 2 Balance the equation from the following two half reactions

H₂C₂O₄ --> 2CO₂ + 2H⁺ + 2e^-
Cr₂O₇^-2 + 14H⁺ + 6e^- --> 2Cr⁺³ + 7H₂O

Solution:
Multiply the first equation by 3 and add them algebraically so the electrons in the two half-reaction equations cancel.

3H₂C₂O₄ --> 6CO₂ + 6H⁺ + 6e^-
Cr₂O₇^-2 + 14H⁺ + 6e^- --> 2Cr⁺³ + 7H₂O

Add the two equations. Cancel the electrons and remove right side H⁺:
Overall equation: 3H₂C₂O₄ + Cr₂O₇^-2 + 8H⁺ --> 6CO₂ + 2Cr⁺³ + 7H₂O

CHEMIX solution:
Insert the two half cell reactions:
H₂C₂O₄ > 2CO₂ + 2H⁺ + 2e^-
Cr₂O₇^-2 + 14H⁺ + 6e^- > 2Cr⁺³ + 7H₂O
and calculate. Example 3 What is the equilibrium constant for the reaction of metallic Cu with bromine to form Cu⁺² and Br^- at 25^oC ?
Br₂ + 2e^- --> 2Br^- E⁰ = 1.09 V
Cu --> Cu⁺² + 2e^- E⁰ = -0.34 V
---------------------------------
Cu + Br₂ --> Cu⁺² + 2Br^-

Solution:
The overall cell voltage : E⁰_cell = 1.09 V -0.34 V = 0.75 V
Calculate

G by inserting n=2 and E⁰_cell=0.75 in eq.:

G = -n F E⁰
Insert

G in eq.: ln K = -

G/RT and calculate.
ln K = 58.36 --> K = e^58.38 = 2.27E25

CHEMIX solution:
Step 1) Insert first and second half-reaction
Br₂ + 2e^- > 2Br^- E⁰ = 1.09
Cu > Cu⁺² + 2e^- E⁰ = -0.34
and calculate equation and overall cell voltage.
Step 2) Insert n=2 (-n=-2) and E⁰_cell=0.75 in equation:

G = -n F E⁰ and calculate.
Step 3) Transfer result of

G to equation: -

G = R T ln K and calculate K.
Example 4 Balance and decide the overall cell potential (E⁰_cell), G and K. : Fe⁺²+ O₂ + H⁺ --> Fe⁺³ + H₂O
knowing that:
Fe⁺³ + e^- --> Fe⁺² E⁰ = 0.77 V
O₂ + 4H⁺ + 4e^- --> 2 H₂O E⁰ = 1.23 V

Solution:
Turn upper half-reaction according to species in unbalanced equation and change sign of E⁰ (0.77 V --> -0.77 V).
&nbspFe⁺² --> Fe⁺³ + e^- Multiply by 4
&nbspO₂ + 4H⁺ + 4e^- --> 2 H₂O
--------------------------------------------
Overall: 4Fe⁺²+ O₂ + 4H⁺ --> 4Fe⁺³ + 2H₂O

The overall cell voltage can be summed from the half-cell potentials of the oxidation and of the reduction reactions.
E⁰_cell = -0.77 V + 1.23 V = 0.46 V
Calculate

G by inserting n=4 (-n=-4) and E⁰_cell=0.46 V in eq.:

G = -n F E⁰
Insert

G in eq.: -

G = RT ln K (ln K = -

G/RT) and calculate.
ln K = 71.62 --> K = e^71.62 = 1.27E31

CHEMIX solution:
Step 1) Insert first and second half-reaction (remember to turn first half-reaction)
Fe⁺³ + e^- > Fe⁺²
O₂ + 4H⁺ + 4e^- > 2 H₂O
and calculate equation and overall cell voltage.
Step 2) Insert n=4 (-n=-4) and E⁰_cell in equation:

G = -n F E⁰ or ln K = -

G/RT and calculate.
Step 3) Transfer result of

G to equation: -

G = R T ln K and calculate K.

Spectroscopy is the study of quantized interaction of energy (typically electromagnetic energy) with matter. The derivation of structural information from spectroscopic data is an important part of chemistry. A common way to simplify this process is to combine information from major spectroscopic and spectrometric techniques as NMR¹,IR² and MS³.
1) NMR - Nuclear Magnetic Resonance Spectroscopy
2) IR - Infrared Spectroscopy
3) MS - Mass Spectrometry

Fig.4 ¹³C-NMR spectra. Off resonance decoupled (upper) and proton decoupled with a ¹³C line in an typical area (209ppm) for C=O.

Mass Spectrometry - Show isotope formula : Show C[12]2H[1]6 (isotope formula) instead of C2H6 .
Text field (upper): Fragment masses separated by spaces
+/-m_e^- : Accuracy for the search/iteration in electron masses (1840 m_e^- = 1 m_n).
Iterate -> : A method in which finds solutions to MS problems by the use of selected isotopes. CHEMIX will try to match input masses by combining masses of all the selected isotopes and their index values (iterative limits). If to many isotope combinations matches the input data, a dotted line (...) will be seen. This problem can be avoided by a reduction of the (+/-m_e-) value. By not selecting Iterate, CHEMIX will perform a search in the MS-data file.
IR and NMR Spectroscopy (H[1] and C[13]) Text fields : IR wave numbers (cm^-1) or NMR ppm values.
Interpretation of parameters and results (1/1), (1/2), (2/3).... (hits/potential hits)
(1/1) = 1 hit of potential 1 hit (Hit% = 100)
(1/2) = 1 hit of potential 2 hits (Hit% = 50)
(2/2) = 2 hits of potential 2 hits (Hit% = 100)
(5/8) = 5 hits of potential 8 hits (Hit% = 5/8*100 = 62.5)

LPHV (Lowest Potential Hit Value) represent the lowest accepted denominator in a fraction (1/LPHV). The result of increasing LPHV from 1 -> 2 (at least two resonance/absorption areas must be present in data file), is that even a 100 hit% as (1/1) will be ignored as a valid hit.
Save Save experiment : MS(+/-m_e^- + selected isotopes + spectral data) + IR/NMR spectral data.
The default settings for Spectroscopy can be changed and saved by the use of Settings -> Save settings

Introduction

Often, we are faced with the task of finding the relationship between two sets of numbers, so that interpolation and extrapolation can be done. Many laboratory experiments are sources for huge amounts of numbers in which, if possible, should be replaced and represented by an "easier to read" function. Measurements of physical properties such as viscosity, density, vapor pressure or thermal conductivity are examples of such data sources. If there is enough data information to find the relationship between X and Y, interpolation and extrapolation is possible. The relationship was, in fact in earlier times, often found by plotting data on various graphs i.e. linear or log-log to see which one gave the best fit.
To help determine how good the fit is for the curve selected, the SSE (Sum of Squares Error) and the coefficient of fit performance (r²) are reported. The best fit usually have the smallest SSE and a coefficient of performance near 1.000. The curve is automatically refitted anytime Calculate push button is used or a change in a curve type occurs.
Fitting functions to empirical data are not always a straight forward procedure. Two functions may have an approximately equal coefficient of performance. If so, as a rule, the simplest function of these should be selected.

Starting a New Data Set

To fit a curve to a new data set, just remove the content (if any) in Data (X Y) edit field and insert your data.
Numeric data must be separated by one or more spaces.
Examples of legal numbers are: -1.234, -1.234E-4, 1.234, 1.234E4
The Curve Fit tool in CHEMIX allows you to define a data set title, assign axis titles and select a curve type. The Edit list box push buttons allows you to save, replace and delete your X Y data sets.

Loading an Existing Data Set

To load a previously saved data set, double click one of the elements in the Name (X Y data) list box. The application reads the data from a file and displays it in the Data (X Y) text field.

Saving Data

Once you have entered a data set, enter a data title in the Name (X Y data) edit field and save your data using one of the push buttons in the Insert frame (Before or After) in Edit list box

Function, r² and SSE

Function: Function in which represent the output curve.
r² The goodness of the fit - 1.0 is perfect.
SSE Sum of Squares Error. The sum of (Y_i-Y_func)².

Function Plot f(x)

The CHEMIX function plotter allows you to insert and plot math functions. Available functions are described in Calculator. Enter a function e.g. sin(x) in the Plot function f(x) text field and Calculate. A plot can now be seen in the plot area.

Generating a New Data Set by a Function g(x)

IMPORTANT: Before you can generate a new data set, you must remove the content (if any) in the Data (X Y) edit field. A simple way to remove the content is by clicking the right mouse button in the Data (X Y) edit field and delete by the use of the mouse-menu.
Now, by inserting a function (say x^2) in the g(x)-text field and proper (X-Max,X-Min) limit values for the function you may Calculate.
The data set will appear both in the (X Y) edit field (as numbers) and as visual points on the screen.

Derivatives

You may plot derivatives of both the inserted function and XY-data simply by selecting one of the radio buttons (Y' or Y'') in the Derivatives area.
NOTE: Taking derivatives of XY-data will not alter any of the data presented in the XY Data field.

Integration - Definite Integrals

Three situations involving one or two functions may occur, these are:
a) int. f(x)dx --> Integration using the function plotter (f(x)).
b) int. g(x)dx --> Integration using data from XY Data field.
c) int. (f(x)-g(x))dx --> Integration using the function plotter (f(x)) and data from XY Data field (g(x)).

CHEMIX will automatically integrate a curve/plot 'on the run'. The result will be presented numerically in front of the integration sign. Simply integrate a function f(x) by inserting a function e.g. X in the f(x) function field and insert proper integration limits in X-max and X-min text fields.

Calculate the area between two functions f(x)=X and g(x)=X^2-2 limited by X=-1 and X=2.

Solution:
1) If not selected, select: Interpolate --> Nat. cubic spline
2) Insert the X-limits -1 and 2 in X-min and X-max fields.
3) Insert X^2-2 in the g(x) text field. Note: The Data X,Y text field must be empty before this operation.
4) Calculate
5) Insert X in the f(x) text field.
6) Calculate
The result of the integration will be presented as a value in the front of the integral sign.

Limitation of Integraton You can not use integration limits that are higher or lower than the X-values presented in the XY Data field when selected interpolation (polylines..).

Note: Calculating integrals of derivatives will lower the accuracy of the integration. This is indicated by a '~'-sign instead of the usual '='-sign.

Implicit multiplication

The CHEMIX Calculator implicit multiplication samples:
(x-2)(x+3) --> (x-2)*(x+3).
pipie2e2E2sin(X)e^2 --> pi*pi*2*e*2E2*sin(x)*e^2

Zoom

Zoom by moving the mouse cursor in plot area while left mouse button is pressed. The zoom actually occurs when releasing the mouse button. If you want to go back to initial XY-max/min values (un-zoom), simply press right or left mouse button in the plot area and release without moving the mouse cursor.
Note: If any zoom exceeds factor 1/2000, initial xy-max/min values (before zoom) will automatically be selected.

Max/Min Limits in a Plot

A built-in feature in CHEMIX uses a set of rules finding the y-limits in a function plot.
If both XY-data and a function are present,- it will be the extreme max/min limits of the xy-data in which will decide these limits. In the case of the presence of function plot only, y-limits will be calculated automatically and x-limits manually.

Radians & Degrees

A function plot may involve trig.func. as sin(x),cos... You may select Radians or Degrees (Degrees unselected).

Data Manipulation g(x,y)

It is possible to manipulate inserted or generated XY-data . This can be done by inserting a function in the Data Manipulation f(x,y) field an Calculate. None of the raw data seen in the Data XY-field will be altered during this operation,.. the result can only be seen graphically as a plot. It is possible to insert both X and Y.
E.g. Y=Y meaning Y_{i (new)}=Y_{i
(old)} does not change anything while Y=Y/X meaning Y_{i (new)}=Y_{i (old)}/X_i does change all the Y-values in the plot. All available functions are described in Calculator.

Interpolate (Data X Y)

By selecting Interpolate, lines may be drawn between points. Polyline draw straight lines between the points. Natural cubic spline assigns third order polynomes to the points.

By clicking the "copy to clipboard" push button (two rectangles) located bottom right, the image will be copied to the clipboard. The image can thereafter be used by other applications (word processors etc.) that has the paste option.

Printing

You may send the plotted curve to a printer by selecting the Print Button (printer graphic). The hard copy will have the same format as seen on the screen.

Calculate

Plot the selected curve by the use of the Calculate push button. This button also automatically calculate max/min limits for the plot, unless the content in one of the X-min X-max Y-min... (manual limits) fields prior to the use of this button has been altered.
NOTE:In some cases when a data set contain "illegal values", as when the denominator in a hyperbolic function equals zero, no fit/plot will occur.

It is possible to change print and copy to clipboard image quality. Simply go to Setup and select print and copy to clipboard image quality by checking/unchecking proper check boxes.

In chemistry Ternary Diagrams are used for depicting chemical compositions - phase diagrams. These diagrams are three-dimensional but is illustrated in two-dimensions for ease of drawing and interpretation. In ternary diagrams the relative percentage (normally weight %) of three components are represented by A, B and C. The only requirement is that the three components have to sum to 100%. If they don't, you have to normalize them to 100%.

Both symbol and text can be inserted in CHEMIX ternary diagrams. A and B can only be inserted as % (e.g.0-100). Text fragments in which may follow the numbers A and B must only partly contain numbers e.g. 10 10 L1 where L1 is the text fragment.

Data Entry Commands: There are seven commands that can be inserted in the end of each text string, these are:
[sf] = small font
[bf] = big font
[-s] = hide symbol
[-t] = hide text
[ih] = ignore hide (will ignore a "global" check box hide if Hide - Text/Symbols- has been selected).
[il] = ignore limits - This is a way to put text or/and symbols outside the triangular frame.
[_???] = Angled text in degees e.g. [_180] (180 deg).

Examples:
20 50 Comment[-t] Symbol only (no text)
30 40 Text Only[-s] Text only (no symbol)
20 50 Text Only[sf] Small Font
30 30 Text Only[bf] Big Font
40 40 Big Text Only[-s][bf] Big Font (no symbol)
20 50 Comment[-t] Symbol only (no text)
40 40 Text Only[-s][-t] Absolutely nothing (no symbol and no text)
20 15 TextOnly [-s][ih] View the text (TextOnly) even if Hide - Text (check box) has been selected
10 10 [180]degrees text[_180][-s] 180 degrees text (no symbol)
-5 -5 Text outside frame[il][-s] Text outside triangular frame (no symbol)

Text Insertion: Insert two numbers (A and B normally weight %) and text representing a phase (e.g. L1) separated by spaces. Because L1 is not a number it will be identified as text in which always must be placed after the two numbers A and B e.g. 20 35 L1 (A=20% , B=35% Text=L1).
Negative numbers are not allowed. C will be calculated as C=100%-A-B. The sum (A+B) must not exceed 100%.

Identify a number as text: A number can be identified as text if it is enclosed in square brackets.

SAMPLES:
20 20 number [10][sf] --> number 10 (small font)
25 25 number[11] --> number 11 (normal font)
30 30 number [12][bf] --> number 12 (big font)

Calculate: After inserting or changing the content in the A(%) B(%) Text-field, you have to re-calculate

Normalization: This is a way to convert e.g. A B C into A% B%. Before normalization make sure that data has been inserted in A B C format. Normalization will change inserted A B C values into A and B percent values.

As we see below these values vill be converted into A% B% format:

A% B%
_____________
12.30 48.66
31.04 45.42
48.24 44.12
15.22 50.00

There are three options regarding the graphical content of a legend box. These are: 1) Symbol 2) Symbol + Line and 3) Line.

Splines: By selecting spline mode you will be able to create splines. Spline points may be created interactively by clicking the mouse pointer (cross) within the triangular/frame area (making small rectangles). First and last point may be connected by selecting connect. You may move spline points by hand. Simply locate the mouse pointer within a small rectangular area, click the left mouse button and move the mouse pointer/rectangle. Spline points may be removed by clicking the right mouse button with the mouse pointer located within a rectangular area. A thick lined rectangle/(spline point) indicates that the spline point belongs to current selected set of spline points (1 spline 2 spline). Also, every set of spline points has a number located on the right side of the first created spline point . This number is the spline set number. 30 independent sets of splines each containing up to 50 spline points may be selected/created from the splines - drop down selection box.

Delete : Removes all spline points for current selected set of spline points.

1 spline 2 spline ... : Sets of spline points. Select one of these from the drop down selection box. Spline sets that are in use has IU (In Use) located on the right side ( 1 spline IU 2 spline IU ...)

Fill mode enables you to insert fill points. First select a color from the drop down selection box. Insert a fill point by clicking left mouse button inside the frame. Fill points can be moved around inside the frame using left mouse button. You can remove fill points (cross) by locating the mouse cross inside a fill-point and right-click it. You can also remove a fill point by the Delete push button.

1 fill 2 fill... : Select one of these (Note: Fill points that are in use has IU (In Use) located on the right side.

Place the mouse cursor over the legend box an click left mouse button. Keep holding left mouse button down and move the box around. Note: The legend box can not be seen before text has been inserted in the legend text field.

Symbols : Uncheck to hide all symbols (Note: It will not hide symbols for entries containing a [ih] (ignore hide) command).

Text : Check to view all x,y entry text (Note: It will not hide text in entries containing a [ih] (ignore hide) command).

Dot : Check to select a dotted line grid - Uncheck to select a solid line grid.

Percent: Check to view percent-scale (0-100) - uncheck to view a fraction scale (0-1)

Copy to Clipboard : By clicking the "copy to clipboard" push button (two triangles) located bottom right, the image will be copied to the clipboard. The image can thereafter be used by other applications (word processors etc.) that has the paste option.

It is possible to change print and copy to clipboard image quality. Simply go to Setup and select print and copy to clipboard image quality by checking/unchecking proper check boxes.

In chemistry, geology, mineralogy and materials science, a phase diagram is a type of graph used to show the equilibrium conditions between the thermodynamically distinct phases. In a pressure-temperature diagram of water (single substance), the axis correspond to the pressure and temperature. In such a phase diagram the lines represent the equilibrium or boundaries between the phases.

Unlike for single substances, alloys do not have specific phase boundaries but rather tend to solidify over a temperature range. In such a phase diagram the elements is present on opposite side of the diagram (100% A or 100% B) . Occasionally there is a mixture of the constituent elements which produces solidification at a single temperature like a pure element (the eutectic point). The eutectic point can be found experimentally by plotting cooling rates over ranges of alloy composition.

Six sets of x,y data can be entered, each set represented by a plot symbol (black circle, red square etc.)

1) Select a radio button located below one of the symbols representing your data set.

Both symbol and text can be inserted in CHEMIX binary diagrams. A and B can be inserted as any real numbers. Text fragments in which may follow the numbers A and B must only partly contain numbers e.g. 10 10 L1 where L1 is the text fragment.

There are seven commands that can be inserted in the end of each text string, these are:
[sf] = small font
[bf] = big font
[-s] = hide symbol
[-t] = hide text
[ih] = ignore hide (will ignore a "global" check box hide if Hide - Text/Symbols- has been selected).
[il] = ignore limits - This is a way to put text or/and symbols outside the rectangular frame.
[_???] = Angled text in degees e.g. [_180] (180 deg).

Examples:
20 50 Comment[-t] Symbol only (no text)
30 40 Text Only[-s] Text only (no symbol)
20 50 Text Only[sf] Text only (small font)
30 30 Text Only[bf] Text only (big font)
40 40 Big Font Only[-s][bf] Big Font Only (no symbol)
20 50 Comment[-t] Symbol only (no text)
40 40 Text Only[-s][-t] Absolutely nothing (no symbol and no text)
20 15 TextOnly [-s][ih] View the text (TextOnly) even if Hide - Text (check box) has been selected
10 10 [180]degrees text[_180][-s] 180 degrees text (no symbol)
-5 -5 Text outside frame[il] Text outside rectangular frame if x-min and y-min limits > -5

Text Insertion: Insert two numbers A and B and text representing a phase (e.g. L1) separated by spaces. Because L1 is not a number it will be identified as text in which always must be placed after the two numbers A and B e.g. 5.2 7.3 L1 (A= 5.2 B=7.3 Text=L1).

Identify a number as text: A number can be identified as text if it is enclosed in square brackets.

SAMPLES:
20 20 number [10][sf] --> number 10 (small font)
25 25 number[11] --> number 11 (normal font)
30 30 number [12][bf] --> number 12 (big font)

Name - Experiment : Name representing current x,y data sets and parameters "Experiment".

For plotting a function, insert a function in the text field located below "Generate plot" push button. Before using the push button, remember to enter min/max-limits for the plot. By selecting one of the items in the drop down list you will aso be able to plot physical properties such as e.g. melting points etc. You can even manipulate physical (empirical) values selected from the drop down list by adding a function to the physical property (e.g . Melting point(C)+Symbol * sin(x) )

Six options are available. First three options are local options. You must select a symbol (radio buttons located above the legend text field) before selecting one of the three first options.

Deactivate(d) : Line drawing/interpolation for selected symbol will be deactivated.

There are three options regarding the graphical content of a legend box. These are: 1) Symbol 2) Symbol + Line and 3) Line.

Splines: By selecting spline mode you will be able to create splines. Spline points may be created interactively by clicking the mouse pointer (cross) within the rectangular/frame area (making small rectangles). First and last point may be connected by selecting connect. You may move spline points by hand. Simply locate the mouse pointer within a small rectangular area, click the left mouse button and move the mouse pointer/rectangle. Spline points may be removed by clicking the right mouse button with the mouse pointer located within a rectangular area. A thick lined rectangle/(spline point) indicates that the spline point belongs to current selected set of spline points (1 spline 2 spline). Also, every set of spline points has a number located on the right side of the first created spline point . This number is the spline set number. 30 independent sets of splines each containing up to 50 spline points may be selected/created from the splines - drop down selection box.

Delete : Removes all spline points for current selected set of spline points.

1 fill 2 fill... : Select one of these (Note: Fill points that are in use has IU (In Use) located on the right side.

Calculate : Use this push button to calculate max/min limits of a graph based on max/min inserted entry values.

+20% : Use this push button to rescale "zoom out" by 20% in regard to Calculate.

Zoom by moving the mouse cursor in plot area while right mouse button is pressed. The zoom actually occurs when releasing the mouse button. If you want to go back to initial zoom values (un-zoom), simply locate the mouse cursor outside (or in) the plot area and press +release left mouse button without moving the mouse cursor.
Note: If any zoom exceeds factor 1/2000, initial x,y values (before zoom) will automatically be selected.

Symbols : Uncheck to hide all symbols (Note: It will not hide symbols for entries containing a [ih] (ignore hide) command).

Text : Check to view all x,y entry text (Note: It will not hide text in entries containing a [ih] (ignore hide) command).

Dot : Check to select a dotted line grid - Uncheck to select a solid line grid.

Copy to Clipboard : By clicking the "copy to clipboard" push button (two rectangles) located bottom right, the image will be copied to the clipboard. The image can thereafter be used by other applications (word processors etc.) that has the paste option.

It is possible to change print and copy to clipboard image quality. Simply go to Setup and select print and copy to clipboard image quality by checking/unchecking proper check boxes.

Is it possible to deduce physical properties of a molecule only by knowing its structure? Fairly good predictions can be done by combining the knowledge of electronegativity and the understanding of molecular symmetry. Predictions of a molecule ability to dissolve in other molecules can be estimated this way. Related to this, the expression "Like dissolves like" means that, in general, nonpolar substances dissolve in nonpolar solvents and polar substances dissolve in polar solvents. But what's polarity?
Polarity
The polarity of a molecule is a result of how atoms are arranged in the molecule and their electronegativity. In a non ionic compound, a "real" molecule, the atoms are connected by covalent bonds in which are more or less polar.

1)If all the bonds in a molecule are nonpolar, the molecule is nonpolar. (e.g. CH4)
2)If all the bonds in a symmetrically arranged molecule has equal polarity, the overall polarity of the molecule will cancel out. (e.g. CCl4)
3) If the bonds in a molecule are not symmetrically arranged, then the arrangement of the bonds determines the polarity of the molecule. In the water molecule the bonds are not symmetrically arranged (104.5deg). As a result, the oxygen end of the molecule is slightly more negative than the hydrogen end, and the molecule is polar.

The Protein Data-Bank format - PDB
The PDB format is a 3-D file format that intentionally was designed for dealing with protein structures. CHEMIX basically use this format for smaller structures limited to less than 6000 atoms. A lot of free PDB download sources are available on the Internet. You can easily build your own 3-D library by downloading from these sources. In CHEMIX these files are stored in the pdbfiles directory.

CHEMIX 3-D Molecular viewer allow you to load *.pdb structures, rotate and size both the molecule and its atoms. It is also possible to peek into the text content of these files (File info).

Size By these push buttons you may change the size of both molecule and atoms.

Statistics displays statistical data

File info displays some of the text content in these files e.g. COMPND and REMARK.

It is possible to change print and copy to clipboard image quality. Simply go to Setup and select print and copy to clipboard image quality by checking/unchecking proper check boxes.

In the gas phase the molecules are so energetic that they drift apart rather than connect to each other by intermolecular forces (hydrogen bonds etc.). The gas laws below are about the behavior of gases under different physical conditions. Boyle's Law Boyle's law is about the relationship between pressure and volume if temperature and the amount of molecules are held constant.

The volume of a fixed mass of gas is inversely proportional to the pressure at a constant temperature.

The relation can also be written as pressure times volume equals a constant: PV=k

An increasing container volume decreases the pressure.
A decreasing container volume increases the pressure.

When the volume of a container decreases, the distance between the gas molecules shrink. As a result of this they bump into each other more often than if they where farther apart. The increased molecular movements push at the walls inside the container and increases the pressure.
Ideal Gas Law The Ideal Gas Law was first written in 1834 by Emil Clapeyron. Following relations can be expressed as constants (k₁, k₂...k₆) representing six different values.

PV= k₁
V/T = k₂
P/T = k₃
V/n = k₄
P/n = k₅
1/nT = 1/k₆

In order to make one equation that contain them all i.e. P,V,T and n, we can multiply them all.

P³V³ / n³T³ = k₁k₂k₃ k₄k₅ / k₆
Taking the cube root we get:
PV/nT = (k₁k₂k₃k₄ k₅ / k₆)^1/3
An expression in which on the right side of the equation can be presented as a single constant - R - the gas constant. Now we have a single equation representing the relationship between pressure(P), volume(V), mole(n) and temperature(T).

The Combined Gas Law can be derived by multiplying Boyle's law by the laws of Charles and Gay-Lussac.

P₁V₁ = P₂V₂
P₁V₁² / T₁ = P₂V₂² / T₂
P₁²V₁² / T₁² = P₂²V₂² / T₂²
By taking the square root of this result we get the combined gas law:
P₁V₁ / T₁ = P₂V₂ / T₂

When the temperature in a gas increases, the gas molecules will become faster and therefore more energetic. We may say that the temperature represent "the average kinetic energy of the particles of a substance". Knowing that two ideal gases with the same amount of molecules occupies the same volume and that the total amount of kinetic energy in these two volumes must be the equal, less massive gases will diffuse more rapidly than more massive gases at equal pressure and temperature. This according to the kinetic energy equation: E_k=1/2mv²

The relative rates at which two gases under identical conditions of temperature and pressure will diffuse vary inversely as the square roots of the molecular masses of the gases.

Assume following temperature conditions for two different gases:
T₁ = T₂
and by this that these two gases has the same kinetic energy (E_k=1/2mv²)
1/2m₁v₁² = 1/2m₂v₂²
Moving v₂ to the left and m₁ to the right side of the equation we get:
v₁²/v₂² = m₂/m₁
Taking the square root we get:
v₁/v₂ = (m₂/m₁)^1/2

If we know the mass/density and the velocity of a gas, also knowing the mass/density or the velocity of a second gas, we should be able to calculate the velocity or mass/density of the second gas.

The CHEMIX units converter allow conversions between most common units. It is easy in use and a valuable tool for students and professionals. Six categories of units are presented in the units converter.
Categories
Following units categories exist:
1) Temperature
2) Pressure Stress-Force/Area
3) Energy-Heat-Work
4) Power
5) Force-Weight
6) Length
7) Mass
NOTE:The first text field (upper left) in each category is the SI unit field.

Enter the value to be converted in a proper text field and press the Enter key on the keyboard. Converted unit values will then appear in all the remaining (and previous empty) text fields for the selected category.

X-stepped calculations
Molecular calculator command e.g. MCALC(CaSO4)
Scientific constants
Implicit multiplication
Definite integrals

The CHEMIX calculator functions basically in the same manner as a real pocket calculator. Enter numbers or functions either by clicking on the buttons or using the keyboard . The calculator has the ability to calculate 400 Y-values if the X-variable is included in the expression e.g. sin(X).
By selecting Stop value you may calculate a range of values starting with an X-value given in the X-start value text field and stopped by an X-value given in the X-stop value text field.
By deselecting Stop value you may calculate a range of values starting with an X-value given in the X-start value text field and incremented (stepped) by the value that has been inserted in the X-step value field.
The result (X,Y-values) of such a calculation may be copied by the use of left and right mouse buttons and pasted into the Data (X Y) field in the Curve Fit and Function Plotter and viewed graphically.
To enter a number in exponential format, enter the mantissa, followed by the E key, then +/- if required, then the exponent e.g.(2.34*10^-2 --> 2.34E-2).

Definite integrals - X Calculations

Intagrals will automatically 'on the run' be calculated if the expression includes the X-variable.

Implicit multiplication

The CHEMIX Calculator recognizes implicit multiplication.
Samples:
(x-2)(x+3) --> (x-2)*(x+3).
pipie2e2E2sin(X)e^2 --> pi*pi*2*e*2E2*sin(x)*e^2

Errors

If you an error occur, the calculator result display will contain an error message.

Calculator

Functions/const./operators Example Interpretation

+ X+2 addition

- X-2 subtraction

* X*2 multiplication

/ X/2 division

^ X^2 power X²

( ) 5*(X+2) braces (grouping)

. 2.536 decimal separator

PI PI constant 3.141..

e or EXP e^x or EXP(X) e = constant 2.718..

E 2.5E-2 2.5*10^-2

X 2*X variable

SIN SIN(X) sine function

COS COS(X) cosine function

TAN TAN(X) tangent function

ASIN ASIN(X) arc sine function (result expressed between -1/2 and 1/2 PI or -90 and 90 Deg)

ACOS ACOS(X) arc cosine function (result expressed between 0 and PI or 0 and 180 Deg)

ATAN ATAN(X) arc tangent function (result expressed between -1/2 and 1/2 or -90 and 90 Deg)

SINH SINH(X) hyperbolic sine of X

COSH COSH(X) hyperbolic cosine of X

TANH TANH(X) hyperbolic tangent of X

RAD RAD(180) Convert from degrees to radians

DEG DEG(PI) Convert from radians to degrees

EXP EXP(X) natural exponential e^x

LN LN(X) natural logarithm

10^X 10^(X) 10-pow. function 10^x

LOG LOG(X) logarithm to base 10

SQRT SQRT(X) square root

ABS ABS(-2.43) absolute value

RND RND(10) random number

MCALC MCALC(CaSO4) Molecular Calculator function (Arg. by ',' not legal)

CLR
clear all text

_ <-DEL
backward delete

Functions/const./operators	Example	Interpretation
+	X+2	addition
-	X-2	subtraction
*	X*2	multiplication
/	X/2	division
^	X^2	power X²
( )	5*(X+2)	braces (grouping)
.	2.536	decimal separator
PI	PI	constant 3.141..
e or EXP	e^x or EXP(X)	e = constant 2.718..
E	2.5E-2	2.5*10^-2
X	2*X	variable
SIN	SIN(X)	sine function
COS	COS(X)	cosine function
TAN	TAN(X)	tangent function
ASIN	ASIN(X)	arc sine function (result expressed between -1/2 and 1/2 PI or -90 and 90 Deg)
ACOS	ACOS(X)	arc cosine function (result expressed between 0 and PI or 0 and 180 Deg)
ATAN	ATAN(X)	arc tangent function (result expressed between -1/2 and 1/2 or -90 and 90 Deg)
SINH	SINH(X)	hyperbolic sine of X
COSH	COSH(X)	hyperbolic cosine of X
TANH	TANH(X)	hyperbolic tangent of X
RAD	RAD(180)	Convert from degrees to radians
DEG	DEG(PI)	Convert from radians to degrees
EXP	EXP(X)	natural exponential e^x
LN	LN(X)	natural logarithm
10^X	10^(X)	10-pow. function 10^x
LOG	LOG(X)	logarithm to base 10
SQRT	SQRT(X)	square root
ABS	ABS(-2.43)	absolute value
RND	RND(10)	random number
MCALC	MCALC(CaSO4)	Molecular Calculator function (Arg. by ',' not legal)
CLR		clear all text
_ <-DEL		backward delete

A chemistry terms dictionary is included in the program. Select a word (list box element) by a double click.

The CHEMIX solubility chart has color and solubility information about typical precipitates forming during inorganic qualitative analysis. The chart may be of great help identifying unknown precipitates.

To get information of a precipitate - simply click the mouse cursor on one of the colors in which represent the color of the precipitate. The solubility information will then be visible in the text field below the chart.

Also included in the solubility chart are flame test reference colors for atoms as Li , Na, K, Rb, Ca and Ba.

Checked (Best quality) = Resolution ratio: (Best quality)/(screen dump quality) = 9/1
Unchecked (Good quality) = Resolution ratio (Good quality)/(screen dump quality) = 4/1

Checked - Thin line (all splines) as default
Unchecked - Thick line (all splines) as default

Checked - Connect (all splines) as default
Unhecked - Spines unconnected (all splines) as default

Checked - Thin line (all splines) as default
Unchecked - Thick line (all splines) as default

Checked - Connect (all splines) as default
Unchecked - Spines unconnected (all splines) as default

By selecting "save settings", the selected radio buttons (e.g. periodic table) will be used as "default" the next time the application is started.

1.1 Periodic Table - Symbols:
What is the symbol for:
Potassium
Hydrogen
Sodium
Calcium
Sulfur
Oxygen
Carbon
Lithium
Manganese
Tin

Solution: Activate radio button Name and compare push button symbols by edit field names.

1.2 Periodic Table - Names
Find names for following symbols:
Br
Sn
Ba
H
Cl
C
P
Al
Si
K
Fe
Pb

Solution: Activate radio button Name and compare edit field names by push button symbols.

1.3 Periodic Table - Atomic number
Find the atomic number of:
Uranium
Lead
Phosphorus
Zinc
Tin
Iodine
Nitrogen
Oxygen
Sodium
Neon
Boron
Chromium

Solution: Activate radio button Atomic number.

1.4 Periodic Table - Acid-base properties (oxides)
Find the properties (acid-amph.-basic) for the oxides of the following elements:
Sulfur
Potassium
Chlorine
Manganese
Fluorine
Aluminium
Iodine
Sodium
Boron
Nickel
Copper
Lithium

Solution: Activate radio button Acid-base properties..

1.5 Periodic Table - Phase/state:
Where can we locate the gases in the periodic table.

Solution: Activate radio button Phase and search for "Gas" in the edit fields.
1.6 Properties - Isotopes
Find the number of stable isotopes and their abundance:
Hydrogen
Carbon
Sodium
Tin

Solution: Select push buttons H, C ... and activate radio button Stable isotopes in "Properties".

1.7 Properties - History
Oxygen: Explain the bright red and yellow-green colors of the Aurora
Hydrogen: How many stable and unstable hydrogen isotopes exist. What are their names ?
Uranium: What uranium isotope causes fission (nuclear energy).
Platinum: What will happen if platinum is inserted in an oxygen-hydrogen atmosphere ?
Nitrogen: Find atmospheric vol.% of nitrogen. Determine formula: ammonia, potassium nitrate and ammonium nitrite.
Carbon: Name three allotropic forms of carbon ?
Helium: Describe the use of liquid helium.

Solution: Select push buttons O, H ... and activate radio button History in "Properties" .

1.8 Explain the relationship/connection: electronegativity/atomic radius:

Solution: Alter between the radio buttons Elecronegativity/Atomic radius.

1.9 Decide max/min values of:
Electronegativity
Atomic radius
Melting point
Boiling point
Electrical conductivity
Thermal conductivity
Number of stable isotopes
Density

Solution: Activate Graphics in the periodic table. Use radio buttons (electronegativity...) and search for max/min values.

2.1 Calculate moles:
a) 12 grams of NaCl
b) 23 grams of CaO
c) 9.43 grams of CuSO₄*5H₂O

Solutions:
a) 12gNaCl
b) 23gCaO
c) 9.43gCuSO4*5H2O

2.2 Calculate grams:
a) 0.1 moles of CuCl₂
b) 0.5 moles of NH₄Cl
c) 2 moles of CaSO₄*5H₂O

Solutions:
a)1mCuCl2
b) 0.5mNH4Cl
c) 2mCaSO4*5H2O

2.3 Calculate moles of carbon (C) in:
a) 3.2 moles of CH₄
b) 0.5 moles of HCN
c) 2.17 moles of K₄Fe(CN)₆
Solutions:
a) 3.2mCH4
b) 0.5mHCN
c) 2.17mK4Fe(CN)6

2.4 Calculate grams of oxygen (O) in:
a) 3 grams of CaO
b) 21 grams of H₂O₂
c) 0.23 grams of BaSO₄

Solutions:
a) 3gCaO
b) 21gH2O2
c) 0.23gBaSO4

2.5 Calculate moles of hydrogen (H) in:
a) 1.4 grams of LiH
b) 0.3 grams of H₂O
c) 12 grams of CH₃CH₂OH

Solutions:
a) 1.4gLiH
b) 0.3gH2O
c) 12gCH3CH2OH

2.6 Calculate grams of Sodium (Na) in:
a) 0.6 moles of NaCl
b) 3.4 moles of Na₂CO₃
c) 2.3 moles of NaH₂PO₄

Solutions:
a) 0.6mNaCl
b) 3.4mNa2CO3
c) 2.3mNaH2PO4

2.7 Calculate % Calcium (Ca) in:
a) 2.1 moles of CaO
b) 0.6 moles of CaCl₂
c) 3.2 moles of Ca₃(PO₄)₂

Solutions:
a) 2.1mCaO
b) 0.6mCaCl2
c) 3.2mCa3(PO4)2

2.8 Calculate % Fluorine (F) in:
a) 1.2 grams of HF
b) 2.7 grams of MgF₂
c) 0.9 grams of AlF₃

Solutions:
a) 1.2gHF
b) 2.7gMgF2
c) 0.9gAlF3

2.9 Calculate:
a) % K, Cr and O in K₂CrO₄.
b) Grams of: Ca, C and O in 0.3 mol CaCO₃.
c) Moles of Pb and O in 12grams PbO₂.

Solutions:
a) K2CrO4.
b) 0.3mCaCO3.
c) 12gPbO2.

2.10 Calculate grams and moles (all elements) in Mg₃(PO₄)₂.

Solution:
Mg3(PO4)2

2.11 Calculate moles:
a) 2 moles of Na in NaCl
b) 0.3 moles of Cl in AlCl₃
c) 3.6 moles of P in Ca₃(PO4)₂
Solutions:
a) 2mNa,NaCl
b) 0.3mCl,AlCl3
c) 3.6mP,Ca3(PO4)2

2.12 Calculate grams of:
a) CaO containing 2 grams of O
b) HCl containing 5 grams of Cl
c) Fe₂O₃ containing 2.4 grams of Fe

Solution:
a) 2gO,CaO
b) 5gCl,HCl
c) 2.4gFe,Fe2O3

2.13 Calculate moles of Potassium (K) in:
a) KNO₃ containing 3.2 moles of O
b) KClO₄ containing 0.2 moles of Cl
c) CaCl₂ containing 1.4 moles of Ca

Solutions:
a) 3.2mO,KNO3
b) 0.2mCl,KClO4
c) 1.4mCa,CaCl2

2.14 Calculate grams of Nitrogen(N) in:
a) NO containing 12 grams of O
b) NH₃ containing 0.5 grams of H
c) NH₄Cl containing 3.0 grams of Cl

Solutions:
a) 12gO,NO
b) 0.5gH,NH3
c) 3.0gCl,NH4Cl

2.15 Calculate moles of barium (Ba) in:
a) BaO containing 5.4 grams of O
b) BaCl₂ containing 3.7 grams of Cl
c) BaCO₃ containing 1.0 grams of C

Solutions:
a) 5.4gO,BaO
b) 3.7gCl,BaCl2
c) 1.0gC,BaCO3 2.16 Calculate grams of Cu in:

2.16 Calculate grams of Cu in:
a) CuS containing 0.6 moles of S
b) CuCl₂ containing 6.4 moles of Cl
c) CuSO₄*5H₂O containing 7.8 moles of O
Solutions:
a) 0.6mS,CuS
b) 6.4mCl,CuCl2
c) 7.8mO,CuSO4*5H2O

2.17 Calculate % bromide (Br) in:
a) NaBr containing 2.1 moles of Na
b) CaBr₂ containing 3.0 moles of Ca
c) AlBr₃ containing 0.2 moles of Al

Solutions:
a) 2.1mNa,NaBr
b) 3.0mCa,CaBr2
c) 0.2mAl,AlBr3

2.18 Calculate % lithium (Li):
a) LiI containing 9.0 grams of I
b) Li₂O containing 2.8 grams of O
c) LiHCO₃ containing 1.0 grams of C

Solutions:
a) 9.0gI,LiI
b) 2.8gO,Li2O
c) 1.0gC,LiHCO3

2.19 Calculate
a)% Na, C and O if Na₂CO₃ contains 4 grams of O.
b)Grams of iron and oxygen in Fe₂O₃ containing 0.8 moles of Fe.
c)Moles of lead, carbon and hydrogen in Pb(C₂H₅)₄ containing 4.0 grams of hydrogen.

Solutions:
a) 4gO,Na2CO3
b) 0.8mFe,Fe2O3
c) 4gH,Pb(C2H5)4

2.20 Calculate %, grams and moles of all elements in C₄H₉OH.

Solution:
C4H9OH

2.21 Calculate max.amount of formed Ca₃(PO₄)₂ using 3 grams of Ca ,10 grams of O and 14 grams of P?

Solution:
Determine by calculating following argumented formulas: 3gCa,Ca3(PO4)2 and 10gO,Ca3(PO4)2 and 14gP,Ca3(PO4)2

2.22 How many moles of Be must be used to make 245 grams BeCl₂?

Solution:
245gBeCl2

2.23 The CuSO₄*5H₂O -compound release water when heated. How much mass will evaporate in this process?

Solution:
mass of (CuSO4*5H2O) - mass of (CuSO4) = mass of 5H2O

2.24 Determine empirical formula:
a) 92.83% lead and 7.67% oxygen.
b) 31.9% potassium, 28.9% chlorine and 39.2% oxygen.

Solution: Vary element composition of formula and investigate the element ratio. It will simplify calculation if one of the elements in the trial formulas are argumented by grams.
a) First argumented trial formula , 92.83gPb,PbO shows a plausible element mass ratio.
b) First argumented trial formula, 31.9gK,KClO indicates to little oxygen. Second argumented trial formula 31.9gK,KClO2 also indicates to little oxygen. Third argumented formula 31.9gK,KClO3 shows a correct element ratio.

2.25 0.3 grams of metallic silver was dissolved. A precipitate (0.399 grams of AgX) formed after adding a unknown substance. Determine X knowing that X=Cl- or X=Br-.

Solution: Determine by the use of argumented trial formulas (0.3g) and compare results: 1) 0.3gAg,AgBr or 2) 0.3gAg,AgCl .

2.26 A solution of dissolved Ca(OH)₂ was added CO₂. A white precipitate formed. The precipitate was filtered, dried and weighed. (2.35 grams). How many grams of Ca was involved in the precipitation process?

Solution:
Ca(OH)2+CO2>CaCO3+H2O
2.35gCaCO3

2.27 2 grams of Al was added HCl:
2Al(s) + 6HCl(aq)=2AlCl₃ + 3H₂
a) How many grams of AlCl₃ was formed in the reaction?
b) How many moles of HCl was used forming AlCl₃?

Solutions:
a) 2gAl,AlCl3
b) moles of AgCl3*3= moles of HCl

2.28 3 grams of Fe reacts with 1.29 grams of O₂ forming a pure iron compound.
a) Decide the empiric formula?

Solution:a)First argumented trial formula, 3gFe,FeO contains to small amount of oxygen. Second argumented trial formula, 3gFe,FeO2 contains to much oxygen. Third argumented trial formula, 3gFe,Fe2O2 contains to small amount of oxygen. At least, the formula 3gFe,Fe2O3 shows a correct iron/oxygen ratio.

3.1 Balance:
a) Cu+O2>CuO
b) Fe+Cl2>FeCl3
c) C+Br2>CBr4

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.

3.2 Balance:
a) Zn+HCl>ZnCl2+H2
a) Al+HCl>AlCl3+H2
c) Sn+HCl>SnCl4+H2

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.

3.3 Alkane-oxygen combustion's generally forms CO2 and water. Balance:
a) CH4+O2>CO2+H2O
b) C2H6+O2>CO2+H2O
c) C3H8+O2>CO2+H2O

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.

3.4 Balance:
a) Na2CO3+HNO3>NaNO3+H2O+CO2
b) KClO3+S+H2O>Cl2+K2SO4+H2SO4
c) FeS2+O2>Fe3O4+SO2
d) Al(OH)3+H2SO4>Al2(SO4)3+H2O
e) KBr+MnO2+H2SO4>Br2+MnBr2+KHSO4+H2O
f) Na3SbS4+HCl>Sb2S5+H2S+NaCl
g) Cu+HNO3>Cu(NO3)2+NO+H2O
h) Ca3P2+H2O>Ca(OH)2+PH3
i) FeSO4+KMnO4+H2SO4>Fe2(SO4)3+MnSO4+K2SO4+H2O

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.

3.5 8 grams of C₅H₁₂reacts with oxygen.
Balance:
C5H12+O2>CO2+H2O and determine:
a) How many grams of CO2 was formed in the reaction?
b) How many grams of H2O was formed in the reaction?
c) How many grams of O2 was used during the reaction?

Solutions:8gC5H12+O2>CO2+H2O Copy argumented bal.expr. into "unbalanced equation" edit field and calculate.

3.6 A piece of metallic iron (3g) was dissolved in conc.HCl (6 grams of 100% HCl). The reaction formed H2 and FeCl2.
Balance: Fe+HCl>FeCl2+H2 and determine
a) amount of formed FeCl2
b) amount of formed H2

Solutions: 3gFe+6gHCl>FeCl2+H2 Copy argumented bal.expr. into "unbalanced equation" edit field and calculate.

3.7 Balance: N2+H2>NH3 and determine how many grams of nitrogen and hydrogen used in a reaction forming 15.0 grams of NH3 .

Solution:N2+H2>15gNH3 Copy argumented bal.expr. into "unbalanced equation" edit field and calculate.

3.8 Balance following heated mixture reaction: As2S3+NaNO3+Na2CO3>Na3AsO4+Na2SO4+NaNO2+CO2
Determine:
a) Amount of As2S3, NaNO3 and Na2CO3 used forming 23.0 grams of Na3AsO4 ?
b) How much CO2 will be formed if the "raw material" for the reaction was 10.0 grams As2S3, 75.0 grams NaNO3 and 37.0 grams of Na2CO3 ?

Solutions:Modify balance expr. and calculate
a) As2S3+NaNO3+Na2CO3>23gNa3AsO4+Na2SO4+NaNO2+CO2
b) 10gAs2S3+75gNaNO3+37gNa2CO3>Na3AsO4+Na2SO4+NaNO2+CO2

3.9 Balance:
Na2CO3+HNO3>NaNO3+H2O+CO2
Determine:
a) How much carbon dioxide will be formed using 13.4 grams of sodium in Na2CO3 ?
b) How much H2O will be formed if Na2CO3 contains 3.4 grams of carbon and HNO3 contains 18.8 grams of nitrogen?
c) How many moles of NaNO3 will be formed if CO2 contains 12.3 grams of carbon.

Solutions:Modify balance expr.:
a)13.4gNa,Na2CO3+HNO3>NaNO3+H2O+CO2
b)3.4gC,Na2CO3+18.8gN,HNO3>NaNO3+H2O+CO2
c)Na2CO3+HNO3>NaNO3+H2O+12,3gC,CO2

3.10 Metallic iron may be produced by heating a mixture of Fe2O3 and carbon.
Balance:
Fe2O3+C>Fe+CO2
and determine moles of:
a) Fe2O3 in a reaction forming 1000 grams of Fe.
b) Fe in a reaction using 200 grams of C.
c) CO2 in a reaction forming 12.8 moles Fe.

Solutions:Modify balance expr.
a) Fe2O3+C>1000gFe+CO2
b) Fe2O3+200gC>Fe+CO2
c) Fe2O3+C>12.8mFe+CO2

3.11 FeS2 and O2 reacts forming Fe3O4 and SO2. Balance:
FeS2+O2>Fe3O4+SO2
Determine the amount of formed SO2 and Fe3O4 using 12.0 grams of Fe in FeS2 and 1.5 moles of O in O2?.

Solution:Modify balance expr.
12gFe,FeS2+1.5mO,O2>Fe3O4+SO2

3.12 KClO3 added HCl forms Cl2
Balance:
KClO3+HCl>H2O+KCl+Cl2
Determine:
a)Amount of KClO3 used in a reaction forming 2 grams of Cl2.
b)A compound (KClO3) contains 4 g oxygen. How much Cl2 will be formed adding 0.04 moles of HCl ?

Solutions:Modify balance expr.
a) KClO3+HCl>H2O+KCl+3gCl2
b) 4gO2,KClO3+0.04mHCl>H2O+KCl+Cl2

3.14 1 kg octane reacts with oxygen: C₈H₁₈ + O₂ > CO₂ +H₂O Determine vol. O₂ (NTP) used in this reaction.

Solution: Modify balance expr. and balance:
Step 1: 1000gC8H18 + O2 > CO2 +H2O
Step 2: Calculate O2 -vol. using V=nRT/p.
(n) moles of oxygen (calculated in Step 1):
(gass const.) R=0.082liter*atm/(Kelvin*mol)
(temp.) T=273.15Kelvin
(pressure) p=1atm.

3.15 10.99 grams of a compund containing C, H and O reacts with O2 forming H2O and 21.0 grams of CO2. What's the empiric formula for the compund?

Solution:Find the answer by observing the effect of varying the values of X,Y and Z in following argumented (21g) balance expression:
CXHYOZ + O2 > 21gCO2 + H2O

4.1 Determine H for the combustion of (C₄H₁₀) and (O₂).

Solution: Insert C4H10(g)+O2(g)>CO2(g)+H2O(l) and calculate

4.2 Determine

H when 2 grams Pb (lead) reacts with oxygen forming PbO?

Solution: Insert 2gPb(s)+O2(g)>PbO(s) and calculate

4.3 Balance the equation: Fe(s)+O₂(g) > Fe₂O₃(s).
a)Determine H Is this reaction exothermic/endothermic?.
b)Determine

S
c)Determine H and S if 2 grams of Fe₂O₃ is formed.

Solutions:
a) and b)Insert Fe(s)+O2(g) > Fe2O3(s) and calculate.
c) Insert the modified equation: Fe(s)+O2(g) > 2gFe2O3(s) and calculate.

5.1 Decide Ksp-expr.:
a) PbF2
b) Ag2CrO4

Solutions:
a) Insert eq. of dissociation: PbF2 > Pb+2 + 2F- and calculate the Ksp-expr.
b) Insert eq. of dissociation: Ag2CrO4 > 2Ag+ + CrO4-2 and calculate the Ksp-expr.

5.2 Determine Ksp Ca(OH)2 if 0.0105 mol dissolves in 1 kg of water.

Solution:
Step 1) Insert equation of dissociation: Ca(OH)2 > Ca+2 + 2OH-
Step 2) Insert 0.0105 in the [Ca+2] field (Mass solvent = 1 kg)
Step 3) Calculate

5.3 How many grams of CdF2 dissolves in 1 kg of water when [F-] = 0.234 M ?

Solution:
Step 1) Insert equation of dissociation: CdF2 > Cd+2 + 2F-
Step 2) Insert 0.234 in the [F- ] field (Mass solvent = 1 kg)
Step 3) Calculate (CdF2)

5.4 How many grams of PbCl2 dissolved in 1 kg of water (Ksp=1.78E-5) when [Pb+2] = 0.01 M.

Solution:
Step 1) Insert equation of dissociation: PbCl2 > Pb+2 + 2Cl-
Step 2) Insert 1.78E-5 in Ksp-field (Mass solvent = 1 kg)
Step 3) Insert 0.01 in [Pb+2] common ion field
Step 4) Calculate

5.5 The solubility of Ag2CrO4 in pure water is 6.54E-5M. Show that the solubility of Ag2CrO4 in a 0.05M AgNO3 sol. is 4.48E-10M.

Solution:
Step 1) Insert equation of dissociation: Ag2CrO4 > 2Ag+ + CrO4-2
Step 2) Insert 6.54E-5 in mol-field (Mass solvent = 1 kg)
Step 3) Calculate solubility product.
Step 4) Insert 0.05 in [Ag+] common ion field
Step 5) Calculate

6.1 Calculate [H3O+] and pH in a 0.05M HCN-solution (Ka=5.85E-10).

Solution:
Step 1) Insert equation of dissociation: HCN + H2O > CN- + H3O+
Step 2) Insert (Ka) 5.85E-10
Step 3) Insert [HCN] = 0.05 M
Step 4) Calculate pH and [H3O+]

6.2 Calculate [H3O+] and pH pH in a 0.05M acetic acid solution (Ka=1.75E-5). Is acetic acid a stronger acid than HCN ?

Solution:
Step 1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
Step 2) Insert (Ka) = 1.75E-5
Step 3) Insert [CH3COOH] = 0.05 M
Step 4) Calculate pH and [H3O+]
Step 5) Compare acid strength (acetic acid and HCN).

6.3 Calculate Ka for a 0.1 M solution of C6H5COOH (benzo acid). ([H3O+] = 0.00248 M)

Solution:
Step 1) Insert equation of dissociation: C6H5COOH + H2O > C6H5COO- + H3O+
Step 2) Insert 0.1 in [C6H5COOH] field and 0.00248 in [H3O+] field
Step 3) Calculate 6.4 Calculate Ka for a 0.5 M solution of HCOOH (pH = 2.02)

6.4 Calculate Ka in 0.5M HCOOH (Equilibrium: pH = 2.02).
Solution:
Step 1) Insert equation of dissociation: HCOOH + H2O > HCOO- + H3O+
Step 2) Insert O.5 in [HCOOH] field and 2.02 in pH field
Step 3) Calculate

6.5 Determine [HF] in a pH = 2 solution (Ka=6.94E-4).

Solution:
Step 1) Insert equation of dissociation: HF + H2O > F- + H3O+
Step 2) Insert (Ka)= 6.94E-4
Step 3) Insert pH = 2
Step 4) Calculate
Step 5) Solution: [HF(Before dissoc.)] = [H3O+] + [HF(After dissoc.)].

6.6 Calculate [OH-] and pH if [NH3] = 0.05 M (Kb=1.8E-5).

Solution:
Step 1) Insert equation of dissociation: NH3 + H2O > NH4+ + OH-
Step 2) Insert (Kb) = 1.8E-5
Step 3) Insert [NH3] = 0.05 M
Step 4) Calculate pH and [OH-]

6.7 Calculate pH in a 0.1M NH4Cl solution (KNH4+=5.6E-10).

Solution:
NH4Cl > NH4+ + Cl-
Step 1) Insert equation of dissociation: NH4+ + H2O > NH3 + H3O+
Step 2) Insert KNH4+ = 5.6E-10
Step 3) Insert [NH4+] = 0.1 M
Step 4) Calculate pH

6.8 Calculate [OH-] and pH in a 1 dm3 [NH3] = 0.2 M (Kb=1.8E-5). Decide pH if 2.675 grams of NH4Cl is added?

Solution:
Step 1) Insert equation of dissociation: NH3 + H2O > NH4+ + OH-
Step 2) Insert (Kb) = 1.8E-5
Step 3) Insert [NH3] = 0.2 M
Step 4) Calculate pH and [OH-]
Step 5) Calculate (convert) 2.675 g NH4Cl to mole using the molecular calculator: 2.675gNH4Cl = 0.05 mol, in 1 dm3 -> 0.05 M
Step 6)Insert Kb, [NH3] and [NH4+] (common ion field) from NH4Cl
Step 7) Calculate pH

7.1 What is the overall equation:
Cr₂O₇^-2 + 14H⁺ + 6e-   --> 2Cr⁺³ + 7H₂O
H₂S --> S + 2H⁺ + 2e^-

Solution:
Insert following equations in the half cells:
Cr2O7-2 + 14H+ + 6e- > 2Cr+3 + 7H2O
H2S > S + 2H+ + 2e-
and calculate.

7.2 What is the overall equation:
Cr --> Cr⁺³ + 3e^-
MnO₄^- + 8H⁺ + e- --> Mn⁺² + 4H₂O

Solution:
Insert following half cells reactions:
Cr > Cr+3 + 3e-
MnO4- + 8H+ + e- > Mn+2 + 4H2O
and calculate.

7.3 Balance: OH^- + ClO^- + S₂O₃^-2 --> Cl^- + SO₄^-2 + H₂O

Solution:
Insert following equation i one of the half cell fields:
OH- + ClO- + S2O3-2 > Cl- + SO4-2 + H2O
and calculate.

7.4 Dropping a piece of sodium in water. What is the overall reaction. Is this a spontaneous reaction, explain?
2H₂O + 2e^- --> H + 2OH^- E⁰ = -0.83 V
Na⁺ + e- --> Na    E⁰ = -2.71 V
Solution:
Insert following half cells reactions and potentials:
2H2O + 2e- > H + 2OH-
Na+ + e- > Na
Remember to turn second half cell reaction.
Calculate.
Spontaneous reaction? -> Investigate sign of overall reaction potential.

7.5 What is the value of the solubility product constant (Ksp) for AgCl?
AgCl + e- --> Ag + Cl- E⁰ = 0.22 V
Ag    --> Ag+ + e- E⁰ = -0.80 V
---------------------------------
AgCl --> Ag+2 + Cl-

Solution:
Step 1) Insert first and second half-reaction
AgCl + e- > Ag + Cl-   E⁰ = 0.22 V
Ag > Ag+ + e-    E⁰ = -0.80 V
and calculate equation and overall cell voltage.
Step 2) Insert n=1 (-n=-1) and E0 cell=-0.58 in equation: G = -n F E0 and calculate.
Step 3) Transfer result of G to equation: -G = R T ln K (remember to change sign of G) and calculate K in which represent the solubility product constant.

8.1 Use following spectral data to determine the structure of C₇H₈O

MS: 39 53 79 90 107
IR: 700 790
H[1]-NMR: 2.25 7.0
Solution:
Insert spectral data in proper text fields and calculate.
Step 1. Interpretation of common results
MS, IR and H[1]-NMR strongly indicates an aromatic.
MS --> disubst. ; IR (more specific) --> 1,3 disubst.
Step 2. Further investigation
C7H8O - C6H4 (disubst) = CH4O (possibly OH and methyl (CH3))
Verified by H[1]-NMR (Ph-CH3 Methyl H-shift) and MS (CH2-Ph-OH).
The result:This is a di-subst aromatic compound with a methyl and OH in 1,3 position.

8.2 Determine the structure of C₉H₁₀O₂

MS: 43 65 91
IR: 690 740 1750
H[1]-NMR: 1.95 5.0 7.28

Solution:
Insert spectral data in proper text fields and calculate.
Step 1. Interpretation of common results
MS, IR and H[1]-NMR strongly indicates an aromatic.
MS --> Ph-CH2 ; IR --> monosubst. benz ; H[1]-NMR --> Aromatic H
What about these oxygens ?:
MS --> CH3-C=O ; IR --> Esters uconj. C=O & C-O ; H[1]-NMR --> C(=O)-O-CH (esters)
It must be an ester.
Step 2. Further investigation
C9H10O2 - phenyl(C6H5) = C3H5O2
By MS --> CH3C=O (MS) and H[1]-NMR --> C(=O)-O-CH , we can conclude that:
C3H5O2 = -CH2-C(=0)-O-CH3
The result:This is a mono-subst. aromatic compound: Ph-CH2-C(=0)-O-CH3

8.3 Use following spectral data to determine the structure of C₅H₈O₃
MS: 29 45
IR: 1720 3000
H[1]-NMR : 11
C[13]-NMR(multiplicity) : 27.96(t) 29.71(q) 37.83(t) 178.26(s) 207.02(s)

Solution:
Insert spectral data in proper text fields and calculate.
Step 1. Interpretation of common results
MS, IR, H[1]-NMR and C[13]-NMR strongly indicate the attachment of a carboxyl group.
A C=O group (ketone) identified by C[13]-NMR (207.07), MS and IR.
Multiplicity: Carbon in both groups = singlets
Step 2. Further investigation
Multiplicity: 27.96(t) = CH2 37.83(t) = CH2 : 29.71(q) = CH3
The result: CH3-C(=O)-CH2-CH2-C(=O)OH

9.1 The following data describes the relationship between Celsius and Kelvin
Celsius , Kelvin
200 , -73.15
300 , 26.85
400 , 126.85
450 , 176.85
500 , 226.85
What's the equation for this relationship ?
Solution:
Insert the data set in the Data: (X Y) text field and calculate. The equation visible in the Function field represent the relationship.

9.2 How many Fahrenheit is 100 oCelsius ?
Celsius , Fahrenheit
-73.15 , -99.67
26.85 , 80.33
126.85 , 260.33
176.85 , 350.33
226.85 , 440.33

Solution:
Insert the data set in the Data: (X Y) text field and calculate. The equation visible in the Function field represent the relationship. Use this equation and calculate Y (X = 100).

9.3 Determine the vapor pressure of butadiene at 0oC knowing following temperature/pressure relationship.
0oC , P (mm Hg)
-61.3 , 40
-55.1 , 60
-46.8 , 100
-33.9 , 200
-19.3 , 400
-4.4 , 760
15.3 , 1520
48 , 3700
76.0 , 7600

Solution:
Insert the data set in the Data: (X Y) text field and select Best Fit . The equation visible in the Function field represent the relationship. Use this equation and calculate Y (X = 0).

9.4 Use following experimental data (time and radiation - counts per second) to determine the half life T1/2 for the Pa[234] isotope.
Time(sec) , radiation (cps)
60 , 66.7
150 , 26.9
240 , 11.0
330 , 4.9
420 , 1.7

Solution:
Insert the data set in the Data: (X Y) text field and select Best Fit. The relationship is represented by the equation Y = A * exp(B*X) visible in the Function field. Use the coeff. B in the generated equation and calculate T1/2 knowing that T1/2 = (ln 2)/B .

9.5 The following data where obtained on the rate of hydrolysis of 16% sucrose in 0.1 mol/L HCl aqueous solutions at 34 oC.
t/min , Sucrose remaining %
9.83 , 96.5
59.60 , 80.2
93.18 , 71.0
142.9 , 59.1
294.8 , 32.8
589.4 , 11.1
What is the order of reaction with respect to sucrose ?

Solution:
Insert the data set in the Data: (X Y) text field and select Best Fit. By the curve, an exact exponential decay, we can conclude that the reaction is of first order.

9.6 Calculate the area between two functions f(x)=X and g(x) = -X limited by X=0 and X=5.
Solution:
1)Insert the X-limits 0 and 5 in X-min and X-max fields.
2) Insert -X in the g(x) text field. Note: The Data X,Y text field must be empty before this operation.
3) Calculate
4)Insert X in the f(x) text field.
5) Calculate
The result of the integration will be presented as a value in front of the integral sign.

10.1 Examine the structure of the CO2 molecule. Is this a polar molecule?
Solution:
Load and examine the structure of CO2 - co2.pdb.
Keywords: Symmetry

10.2 Examine the structure of the benzene molecule. Is this a polar molecule? Has benzene a low or high boiling point?

Solution:
Load and examine the structure of benzene - benzene.pdb.
Keywords: Symmetry

10.3 Both benzene and octane contains carbons and hydrogens. What will you expect of the boiling point of a molecule as octane (chain) relatively to a cyclic molecule as benzene.

Solution:
Load and examine the structures of octane and benzene - octane.pdb and benzene.pdb

10.4 What's the reason for the orientation of the H2O molecules in ice.

Solution:
Examine the molecular orientation in ice - ice.pdb.
Keywords: Polarity, electronegativity, lone pair orbitals, temperature and molecular movements.

10.5 What's d the distance between the atoms in diamond and graphite. What's the distance between the layers in graphite

Solution:
Load and examine the structure of diamond and graphite - diamond.pdb and graphite.pdb
By CHEMIX Dictionary - constants find distances.

10.6 What's the structure of sulfur at boiling point?

Solution:
First, load and examine the structure of sulfur - sulfur8.pdb
At boiling point the sulfur atoms will form S8 rings (single bonds). By increasing this temperature further the sulfur atoms will be broken down to S6, S4 and S2 fragments.

10.7 By word explain the structure of DNA. What kind of atoms are represented and what kind of intermolecular bonds exist in DNA.

Solution:
First, load and examine the structure of a DNA fragment - dna.pdb
It seems like a double helix, and the atoms represented are listed in Statistics text field.

11.1 How many moles of gas are found in a 1000 dm3 container if the conditions inside the container are 298.15K and 2 atm?
Solution: Insert following values in proper fields: P = 2 atm, T = 298.15 K, V = 1000 dm3 and press Enter.

11.2 What volume will 120 grams of chlorine gas occupy at STP?

Solution:
First, find moles(n) by the use the molecular calculator and argument the Cl2 by 120g --> 120gCl2
Second, insert the STP values in proper fields (P,n,T) and press Enter.

11.3 A steel tank contains 15.0 g of Cl2 gas under a pressure of 5.0 atm at 22.0 oC. What is the volume of the tank?

Solution:
First, use the units converter to convert 22 oC to Kelvin.
Second, find (n) by inserting 15gCl2 in the molecular calculator and calculate.
Third, calculate V by inserting the values of P ,n and T in the Ideal Gas Law Calculator.

11.4 A balloon (100 g) was at sea level (1atm and 290K) filled with 1000 dm3 of hydrogen gas. Knowing that air contain approx. 21% oxygen, 78% nitrogen and 1% argon, how many grams was this balloon able to lift?.

Solution:
1) Calculate the mass of 1000 dm3 air:
78% of 1000 dm3 =780 dm3 of N2 gas.
Insert following values in the Ideal Gas Calculator:
V=790 dm3, P=1atm, T=290K and press Enter.
multiply the result (n) by 28g/mol and find the N2 mass in the balloon.
Use the same procedure as over, but this time for calculating the masses of O2 and argon.
2) The mass of 1000 dm3 air is obtained by adding mN2, mO2 and argon.
3)Calculate the mass of 1000 dm3 H2 (V=1000 ,P=1atm T=290K) Multiply the result (n) in this calculation by 2g/mol.
4) Result: mN2 + mO2 + mAr - 100g - mH2 = lift

11.5 A gas was confined in a cylinder fitted with a movable piston. At 290K, the gas occupied a volume of 8.0 dm3 under a pressure of 1.85 atm. The gas was simultaneously heated and compressed, so that its volume was 6.45 dm3 and its temperature was 350K. What pressure was exerted by the hot compressed gas?

Solution:
Insert following values in proper fields: T1 = 290 K, V1 = 8.00 dm3, P1 = 1.85 atm, T2 = 350 K, V2 = 6.45 dm3 and press Enter.
Graham's Law of Diffusion
11.6 What is the ratio of the velocity of helium atoms to the velocity of radon atoms (v1 to v2) when both gases are at the same temperature? Mass: Radon=222 u, Helium=4.00 u

Solution:
Insert 222 in the m2 field, 4 in the m1 field and 1 in the v2 field. The v1/v2 ratio will be calculated by pressing Enter.

12.1 Convert 298 oCelsius and 212 Fahrenheit into Kelvin:
Solution:
Insert given temperature values into proper fields and press Enter.

12.2 Convert the following pressures into units of N/m2:
2 atm, 600 mmHg, 55.21 bar, 30 Torr, 4.8 Pa

Solution:
Insert given values into proper fields and press Enter.
12.3 Convert the following energy-heat units into J (Joule):
23.88 cal, 1 Btu, 0.1 kWh, 1 thermie, 10000000 erg

Solution:
Insert given values into proper fields and press Enter.

13.1 Six mixtures consisting of chloroform and water was made in following proportions (CH3Cl/H2O):
Mix.1 = 94/6 , Mix.2 = 75/25 , Mix.3 = 60/40 , Mix.4 = 40/60 , Mix.5 = 25/75 and Mix.6 = 10/90.
Describe the changes that occur when fixed amounts of acetic acid are added 5 times to each of the mixtures.
Insert the results in a phase diagram.
Separate 1-phase area and 2-phase area by splines.

Solution:
Because the relative proportions of water(B)/chloroform(C) remains constant during the addition, the test points will lie in straight lines focusing on A100%.

A B C phase diagram: A = Acetic acid B = H2O C = Chloroform (CH3Cl) (phase = 1-p or 2-p)

Mix. B/C=6/94 Mix. B/C=25/75 Mix. B/C=40/60   Mix.B/C=60/40 Mix.B/C=75/25 Mix.B/C=90/10
A   B   C phase A   B C phase   A B C phase A   B C phase A   B C phase A   B C phase
0   6 94 2-p    0    25 75 2-p 0 40 60   2-p 0 60 40   2-p 0 75 25 2-p 0 90 10   1-p
10 5.4 84.6 1-p   15 21.25 63.75 2-p 20 32 48 2-p 10 54 36 2-p 20   60 20 2-p 10 81 9   1-p
20 4.8 75.2 1-p 30 17.5    52.5   2-p 40   24   36 1-p 20 48 32   2-p 40 45 15   1-p 20 72 8 1-p
30 4.2 65.8 1-p   45 13.75 41.25 1-p 60   16 24 1-p 30 42 28 2-p 60   30 10 1-p 30 63 7 1-p
40 3.6 56.4 1-p   60 10 30 1-p   80    8 12 1-p 40 36 24 1-p 80   15 5    1-p 40 54 6 1-p

13.2 Normalize following values:

Solution: Insert values of A, B and C into the A% B% Text edit field and Normalize.

How many mL of 1-propanol (100%) do we need to prepare 1000 mL of a 2 M 1-propanol (C3H7OH) (aq)solution.
Density = 0.8034g/mL FW=60.0959

Solution:
Calculate the volume of 2 moles of 1-propanol

(C3H7OH) = n*FW / d = ( 2mol *60.0959 g/mol) / 0.8034 g/ml = 149.604 mL

Dilution by volume:
In a flask containing approx. 800 mL of water measure out 149.49 mL of 1-propanol and dilute by adding water until the volume of the solution = 1000 mL

CHEMIX School solution:

Select propanol-1
Insert 100 (100%) and select %w/w (Stock concentration)
Insert V= 1000
Insert M= 2

******** Dilution by volume ********
Put about 600 mL of water into a flask and add 149.604 mL propanol-1 stock conc .Add water until 1000 mL
******** Dilution by mass ********
Put about 600 g of water into a flask and add 120.192 g of propanol-1 stock conc. Add water until 980.083 g

How many mL of a 3 M sodium chloride solution must be used to make 200 mL of a 0.75 M solution

Solution:

C₁V₁ = C₂V₂ C₁=3M, V₁=? , C₂=0.75M , V₂=200ml

V1 = C2*V2 / C1 = ( 0.75M * 200mL)/3M = 50mL

Measure out 50 mL of the start reagent and dilute by adding water until the volume of the solution = 200 mL

CHEMIX School solution:

1 Select substance = sodium chloride
2 Insert Stock concentration = 3 M
3 Insert Concentration M= 0.75

******** Dilution by volume ********
Put about 100 mL of water into a flask and add 50 mL sodium chloride stock conc.Add water until 200 mL
******** Dilution by mass ********
Put about 100 g of water into a flask and add 55.7081 g of sodium chloride stock conc.Add water until 205.743 g

A solution contain 80 grams of NaCl / L . How much of this solution do we need to prepare 1000 mL of a 1 M NaCl solution

Solution:

C(M) =    n/V      =      (80g /58.44 g/mol)/1L        = 1.37 M

C₁ = 1.37M   ,   V₁=? ,    C₂ = 1M    ,    V₂ = 1000 mL

V₁(mL) = V₂ * C₂    / C₁      =      1000 mL * 1M / 1.37M   = 730.53 mL

CHEMIX School solution:

1 Select substance = sodium chloride
2 Insert Stock concentration = 8 (w/v%)
3 Insert Concentration M= 1.0

******** Dilution by volume ********
Put about 200 mL of water into a flask and add 730.531 mL sodium chloride stock conc. Add water until 1000 mL
******** Dilution by mass ********
Put about 200 g of water into a flask and add 769.209 g of sodium chloride stock conc.Add water until 1038.61 g

a)What is the molarity of a 16m (molal) hydrocloric acid ( HCl ) solution(Density=1.18298)
b) What is the percent composition (w/w%)
c) How many ml of a 16m HCl solution is needed for making a
200 ml 4 M solution?

Solutions:
a)
n=moles,m=mass,FW = Formula w., d= density,V=volume,C=conc.

If the mass of the solution is eg. 1000 grams:

C_m=n/m(solv) = (x/FW)/(m(solv)-x/1000) =16m

x=Cm/(Cm/1000+1/FW)

x=16/(16/1000+1/36.4606) = 368.4355g HCl = 10.105 mol

V=1000g/1.18298g/mL=845.32mL

C_M=10.105mol/0.84532L=11.954M

b) Percent composition (w/w%)

%(HCl)=100%*n(HCl)*FW(HCl)/(n(HCl)*FW(HCl)+m(H2O))

= 100%*368.4355g/1000g = 36.8436 %

c) Dilution

V1*C1 = V2 * C2

V1= 200 ml , C1 = 4M , C2 = 12.025M , V2 = ?

V2 = V1*C1/C2 = 200ml*4M/12.025M

= 66.53 mL HCl solu.

a) and b) CHEMIX School solution:

1 Select substance = hydrochloric acid
2 Insert Stock concentration = 16 (molal)

c) CHEMIX School solution:

1 Select substance = hydrochloric acid
2 Insert Stock concentration = 16 (molal)
3 Insert Amount - Vsolu. = 200 (mL)
4 Insert Concentration Molar = 4.0

******** Dilution by volume ********
Put about 100 mL of water into a flask and add 66.9231 mL hydrochloric acid stock conc. Add water until 200 mL
******** Dilution by mass ********
Put about 100 g of water into a flask and add 79.1685 g of hydrochloric acid stock conc. Add water until 213.179 g

After registering you will receive an email containing a License ID and a download URL for a CHEMIX version in which allow a license to be installed. You must install both License ID and License Owner before you can use CHEMIX.

Step 1) Download CHEMIX using the URL given in the received email.
Step 2) Unzip the received ZIP-file.
Step 3)Activate the CHEMIX License installation program CMIXREG.EXE
Step 4) Insert your License ID and License Owner in proper text fields.
Step 5) Use the Install License push button.
If a SUCCESS, your license has been installed and you can start using CHEMIX School program (CHEMIX.EXE) immediately.

Mass (g)	<------>	n (mol)
Volume (mL)	<------>	mass solution (m_s)

Name	Symbol	Units
Molarity	M	(Moles of solute)/(Liters of solution)
molality	m	(Moles of solute)/(Kg of solvent)
Weight%	w/w%	(g of solute)/(100g of solution)
Weight to Volume%	w/v%	(g of solute)/(100mL of solution)
Mole fraction	x(A)	(Moles(A)/(Moles(A)+Moles(B)....)

Introduction

Starting a New Data Set

Loading an Existing Data Set

Saving Data

Function, r2 and SSE

Function Plot f(x)

Generating a New Data Set by a Function g(x)

Derivatives

Integration - Definite Integrals

Implicit multiplication

Zoom

Max/Min Limits in a Plot

Radians & Degrees

Data Manipulation g(x,y)

Interpolate (Data X Y)

Printing

Calculate

X-stepped calculations Molecular calculator command e.g. MCALC(CaSO4) Scientific constants Implicit multiplication Definite integrals

Definite integrals - X Calculations

Implicit multiplication

Errors

Function, r² and SSE

X-stepped calculations

Molecular calculator command e.g. MCALC(CaSO4)

Scientific constants

Implicit multiplication

Definite integrals