ENGTECH 1CH3 Chemistry Assignment Sample McMaster University Canada

ENGTECH 1CH3 Chemistry Test Papers cover the basic principles of chemistry, including atoms and molecules, chemical bonding, states of matter, solutions, and acids and bases. It would also introduce students to more advanced topics such as organic chemistry, biochemistry, and thermodynamics. Students would gain an understanding of the role chemistry plays in our everyday lives.

It provides students with a basic understanding of the building blocks of our world, how they interact with one another, and how they can be manipulated to create new substances. In addition, the ENGTECH 1CH3 Chemistry Assignment Sample course teaches critical thinking skills and problem-solving techniques that are essential for success in any field.

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Assignment Activity 1: To review and apply the concepts of chemical symbols, formulae, atoms and molecules, the mole (mol), and stoichiometry of chemical compounds and chemical composition.

• Chemical symbols are used to represent the elements on the periodic table. The element’s symbol is typically written before the chemical name of the compound. For example, water is represented by H 2 O.
• A chemical formula is a way of representing a compound using symbols. For example, water has the chemical formula H2O – meaning that it consists of two atoms of hydrogen (H) and one atom of oxygen (O).
• An atom is the basic unit of matter, and all elements are composed of atoms. The number of protons in an atom’s nucleus determines what element it is. For example, all atoms of hydrogen have one proton in their nucleus, while atoms of helium have two protons.
• A molecule is a group of two or more atoms held together by chemical bonds. For example, water molecules are composed of two hydrogen atoms and one oxygen atom bonded together.
• The mole (mol) is a unit used to measure amounts of substances. One mole of a substance is equal to its atomic or molecular weight in grams. For example, the atomic weight of hydrogen is 1.0, so one mole of hydrogen atoms would weigh 1.0 grams.
• Stoichiometry is the study of the relative proportions of reactants and products in chemical reactions. In a stoichiometric reaction, all of the reactants are completely consumed and all of the products are formed.

Now let’s put it all together. When we combine two or more elements, they form a compound. The relative proportions of the atoms in a compound are always the same. For example, water (H2O) always has two atoms of hydrogen for every one atom of oxygen, no matter how much water is present.

The relative proportions of the atoms in a compound can be determined by its chemical formula. The formula for water (H2O) tells us that there are two atoms of hydrogen (H) for every one atom of oxygen (O). Therefore, we know that there are twice as many hydrogen atoms as oxygen atoms in a water molecule.

Assignment Activity 2: To discuss the most important rules and concepts for writing chemical reactions as standard equations, calculate reacting amounts and yields of products, and determine theoretically and percentage yields using reactions of industrial importance.

There are several rules and concepts that are important for writing chemical reactions as standard equations. First, the reactants must be written on the left side of the arrow and the products must be written on the right side. Second, the number of each type of atom must be equal on both sides of the arrow. This is called the law of conservation of mass. Third, the overall charge must be balanced on both sides of the arrow. This is called the law of conservation of charge.

To calculate the reacting amounts and yields of products, we need to know the chemical formula for each reactant and product. The yield is the amount of product that is formed from a given amount of reactant. Theoretical yield is the maximum amount of product that can be formed from a given amount of reactant. Percent yield is the actual yield divided by the theoretical yield, multiplied by 100%.

For example, let’s say we have a reaction where we start with 2 moles of reactant A and 3 moles of reactant B. We know that the theoretical yield of product C is 2 moles. However, if the actual yield is only 1 mole, then the percent yield would be 50%. This means that only half of the product C was actually formed.

To determine the theoretical and percentage yields, we need to know the stoichiometric ratios of the reactants and products. The stoichiometric ratio is the mole ratio of the reactants and products in a chemical reaction. For example, in the reaction A + B → C, the stoichiometric ratio of A to B is 1:1 and the stoichiometric ratio of B to C is 3:2.

Now let’s put it all together. In a chemical reaction, the reactants are converted into products. The relative proportions of the atoms in the reactants and products determine the stoichiometric ratio. The stoichiometric ratio can be used to calculate the theoretical yield of a product. The actual yield may be different from the theoretical yield, and the percent yield can be calculated by dividing the actual yield by the theoretical yield.

Assignment Activity 3: To name chemical compounds using different systems in current use and to demonstrate a basic vocabulary of compounds by name and formula.

There are several different systems in current use for naming chemical compounds. The most common system is the International Union of Pure and Applied Chemistry (IUPAC) system. In the IUPAC system, compounds are named by their component parts. For example, the compound water would be named “hydrogen oxide” because it is made up of two parts, hydrogen and oxygen.

Other common systems for naming chemical compounds include the Common system and the Stock system. In the Common system, compounds are named after their common names. For example, water would be named “water” and hydrogen peroxide would be named “hydrogen peroxide”. In the Stock system, compounds are named after the person who first synthesized them. For example, water would be named “H2O” and hydrogen peroxide would be named “HOCl”.

To demonstrate a basic vocabulary of compounds by name and formula, let’s take a look at some common compounds and their formulas. Water is H2O, hydrogen peroxide is H2O2, and vinegar is CH3COOH.

In summary, there are several different systems in current use for naming chemical compounds. The most common system is the IUPAC system, which names compounds after their component parts. Other common systems include the Common system and the Stock system. Compounds can also be identified by their formulas.

Assignment Activity 4: To define appropriate terms and make calculations in the use of chemicals in solutions, including units, solution preparation and solution reaction stoichiometry.

A solution is a mixture of two or more substances in which the molecules of the substances are evenly distributed. The substance that is doing the dissolving is called the solvent, and the substance that is being dissolved is called the solute.

For example, when you dissolve sugar in water, the sugar molecules are evenly distributed throughout the water. The water is the solvent and the sugar is the solute.

The amount of a substance that can be dissolved in a given amount of solvent is called solubility. The solubility of a substance is usually expressed as a concentration, which is the amount of solute per unit volume of solvent.

To calculate the concentration of a solution, we need to know the amount of solute and the volume of the solution. For example, if we have a sugar solution that is made up of 1 gram of sugar dissolved in 100 milliliters of water, the concentration of the solution would be 1 gram per 100 milliliters, or 1% (w/v).

When two solutions are mixed together, the process is called dilution. Dilution can be used to increase the volume of a solution without changing the concentration. For example, if we have a 1% (w/v) sugar solution and we add 100 milliliters of water to it, the concentration of the resulting solution will still be 1% (w/v).

To calculate the concentration of a solution after dilution, we need to know the volume of the original solution, the volume of the added solvent, and the volume of the final solution. For example, if we have a 1% (w/v) sugar solution that is made up of 100 milliliters of water and we add 100 milliliters of water to it, the volume of the final solution will be 200 milliliters. The concentration of the final solution will be 1% (w/v).

Assignment Activity 5: To define basic terms and explain the importance of Arrhenius and Bronsted acids and bases and the pH/pOH concept, and to perform calculations involving acid-base stoichiometry and simple titrations.

An acid is a substance that increases the concentration of hydrogen ions in a solution. A base is a substance that decreases the concentration of hydrogen ions in a solution. The pH of a solution is a measure of the concentration of hydrogen ions in the solution.

The Arrhenius definition of an acid is a substance that increases the concentration of hydrogen ions in a solution. The Arrhenius definition of a base is a substance that decreases the concentration of hydrogen ions in a solution.

The Bronsted-Lowry definition of an acid is a substance that donates hydrogen ions to a solution. The Bronsted-Lowry definition of a base is a substance that accepts hydrogen ions from a solution.

The pH of a solution is a measure of the concentration of hydrogen ions in the solution. The pOH of a solution is a measure of the concentration of hydroxide ions in the solution.

To calculate the pH of a solution, we need to know the concentration of hydrogen ions in the solution. For example, if we have a solution that is made up of 1 gram of hydrogen ions and 100 milliliters of water, the concentration of the solution would be 1 gram per 100 milliliters, or 1% (w/v). The pH of the solution would be 2.

To calculate the pOH of a solution, we need to know the concentration of hydroxide ions in the solution. For example, if we have a solution that is made up of 1 gram of hydroxide ions and 100 milliliters of water, the concentration of the solution would be 1 gram per 100 milliliters, or 1% (w/v). The pOH of the solution would be 14.

An acid-base titration is a procedure used to determine the concentration of an acid or a base. The most common type of acid-base titration is the neutralization reaction between an acid and a base. In a neutralization reaction, the acid and the base react to form water and salt.

To perform an acid-base titration, we need to know the volume of the acid or base solution, the concentration of the acid or base solution, and the volume of the titrant. The volume of the titrant is the volume of acid or base solution that is required to neutralize the acid or base. For example, if we have a 1 molar (M) acid solution and we add 1 molar (M) of a base to it, the volume of the titrant would be 1 liter. The concentration of the final solution would be 1 M.

Assignment Activity 6: To apply the Ideal Gas Law equation, Avogadro’s Law, and Dalton’s Law to mass relationships and stoichiometry where the product is a gas or a mixture of gases.

The Ideal Gas Law equation is PV = NRT, where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the universal gas constant, and T is the temperature of the gas.

Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Dalton’s Law states that the total pressure of a gas is equal to the sum of the partial pressures of the gases that make up the mixture.

To apply the Ideal Gas Law equation to stoichiometry, we need to know the molar mass of the gas. For example, if we have a 1-liter container of nitrogen gas at 1-atmosphere pressure and 25 degrees Celsius, the molar mass of nitrogen is 14.0 grams per mole. The number of moles of nitrogen gas in the container is 1 mole.

To apply Avogadro’s Law to stoichiometry, we need to know the molar mass of the gas. For example, if we have a 1-liter container of nitrogen gas at 1-atmosphere pressure and 25 degrees Celsius, the molar mass of nitrogen is 14.0 grams per mole. The number of molecules of nitrogen gas in the container is 6.02 x 10^23 molecules.

To apply Dalton’s Law to stoichiometry, we need to know the partial pressures of the gases that make up the mixture. For example, if we have a 1-liter container of nitrogen gas at 1-atmosphere pressure and 25 degrees Celsius, the partial pressure of nitrogen is 1 atmosphere. The partial pressure of oxygen in the same container would be 0.21 atmospheres. The total pressure of the gas mixture is 1.21 atmospheres.

When we know the molar mass of a gas, we can use the Ideal Gas Law equation to calculate the number of moles of gas in a given volume. We can then use Avogadro’s Law to calculate the number of molecules of gas in the same volume. Finally, we can use Dalton’s Law to calculate the partial pressure of each gas in the mixture.

Assignment Activity 7: To explain reversible reactions, equilibrium constants, and Le Chatelier’s Principle and apply them to a variety of chemical reactions.

A reversible reaction is a chemical reaction that can proceed in both forward and reverse directions. The forward reaction is the reaction that is being studied, while the reverse reaction is the reaction that would occur if the reactants were reversed.

An equilibrium constant is a value that describes the relative abundance of reactants and products in a chemical reaction at equilibrium.

Le Chatelier’s Principle states that if a chemical system is at equilibrium and is subjected to a change in conditions, the system will adjust itself so as to minimize the effect of the change.

To apply Le Chatelier’s Principle to a reversible reaction, we need to know the equilibrium constant for the reaction. For example, if we have a reversible reaction with an equilibrium constant of 1, and we increase the temperature of the system, the reaction will shift to the side that has the higher energy. In this case, the reactants will have more energy than the products, so the reaction will shift to the left.

Assignment Activity 8: To define the basic terms associated with thermochemistry and thermodynamics including specific heat capacity and Hess’ Law, to apply Hess’ Law to calculate heat changes associated with a variety of chemical reactions, and to explain the First and Second Laws of Thermodynamics and the Gibbs Free Energy term.

Thermochemistry is the study of the heat changes that occur during chemical reactions.

Thermodynamics is the study of the relationship between heat and work.

• The First Law of Thermodynamics states that energy can neither be created nor destroyed, but it can be converted from one form to another.
• The Second Law of Thermodynamics states that the entropy of the universe always increases over time.

The Gibbs Free Energy is a measure of the amount of energy available to do work.

A chemical reaction is exothermic if it releases heat to the surroundings, and endothermic if it absorbs heat from the surroundings.

The specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.

Hess’ Law states that the heat change for a reaction is the same whether the reaction occurs in one step or several steps.

To apply Hess’ Law to a chemical reaction, we need to know the heat change for each step of the reaction. For example, if we have a reaction that occurs in two steps, and the heat change for the first step is -50 kJ/mol and the heat change for the second step is +30 kJ/mol, the overall heat change for the reaction will be -20 kJ/mol.

Assignment Activity 9: To define oxidation numbers and assign them by using simple rules and to balance a variety of oxidation-reduction (redox) reactions.

An oxidation number is a measure of the degree of oxidation or reduction in an atom.

Oxidation numbers can be assigned to atoms by using simple rules:

• The oxidation number of an element in its natural state is 0.
• The oxidation number of a monatomic ion is equal to the charge on the ion.
• The oxidation number of a compound is the sum of the oxidation numbers of the atoms in the compound.
• The oxidation number of a polyatomic ion is the sum of the oxidation numbers of the atoms in the ion.
• The oxidation number of hydrogen is +1, except when it is bonded to a metal, in which case it is -1.
• The oxidation number of oxygen is -2, except when it is bonded to a fluorine atom, in which case it is -1.

To balance a redox reaction, we need to make sure that the oxidation numbers of the atoms are balanced on both sides of the equation. For example, if we have the reaction: 2H + 2e → H2

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