Chemistry CIA #4 Practice Test Notes

Periodic Table of Elements

The periodic table organizes elements by their atomic number and recurring chemical properties. Key elements mentioned include Hydrogen (H), Lithium (Li), Beryllium (Be), Boron (B), Carbon (C), Nitrogen (N), Oxygen (O), Sodium (Na), Magnesium (Mg), Aluminum (Al), Silicon (Si), Phosphorus (P), Sulfur (S), Potassium (K), Calcium (Ca), Scandium (Sc), Titanium (Ti), Vanadium (V), Chromium (Cr), Manganese (Mn), Iron (Fe), Cobalt (Co), Nickel (Ni), Copper (Cu), Zinc (Zn), Gallium (Ga), Germanium (Ge), Arsenic (As), Selenium (Se), Rubidium (Rb), Strontium (Sr), Yttrium (Y), Zirconium (Zr), Niobium (Nb), Molybdenum (Mo), Technetium (Tc), Ruthenium (Ru), Rhodium (Rh), Palladium (Pd), Silver (Ag), Cadmium (Cd), Indium (In), Tin (Sn), Antimony (Sb), Tellurium (Te), Cesium (Cs), Barium (Ba), Hafnium (Hf), Tantalum (Ta), Tungsten (W), Rhenium (Re), Osmium (Os), Iridium (Ir), Platinum (Pt), Gold (Au), Mercury (Hg), Thallium (Tl), Lead (Pb), Bismuth (Bi).

Item 1: Endothermic vs. Exothermic Reactions

The question requires explaining whether a given reaction is endothermic or exothermic.
Exothermic Reaction: A reaction that releases energy (heat) into the surroundings.
Endothermic Reaction: A reaction that absorbs energy (heat) from the surroundings.

Item 2: Molarity Calculations

The question requires calculating the molarity of solutions.
Molarity (M) is defined as moles of solute per liter of solution: $M = \frac{mol}{L}$

Item 2a

A 3.0 L solution contains 0.8 moles of HCl.
Molarity: $M = \frac{0.8 \, mol}{3.0 \, L} = 0.267 \, M$

Item 2b

A 200 mL solution contains 22 g of dissolved $CO_2$ .
Convert mL to L: 200 mL = 0.2 L
Convert grams of $CO_2$ to moles using its molar mass (approximately 44 g/mol).
Moles of $CO_2$ : $n = \frac{22 \, g}{44 \, g/mol} = 0.5 \, mol$
Molarity: $M = \frac{0.5 \, mol}{0.2 \, L} = 2.5 \, M$

Item 3: Ranking Substances by Boiling Point and IMFs

The question involves ranking substances with hydrogen bonded to unknown halogens (W-Z) by boiling point and intermolecular forces (IMFs).
Boiling Point: The temperature at which a substance changes from liquid to gas.
Intermolecular Forces (IMFs): Attractive forces between molecules (e.g., hydrogen bonding).
Substances with stronger IMFs generally have higher boiling points.

Item 4: pH Scale and Soil Acidity/Basicity

The question involves classifying soil samples based on their pH values.
pH Scale: A scale from 0 to 14 that measures the acidity or basicity of a solution.
- pH < 7: Acidic
- pH = 7: Neutral
- pH > 7: Basic

Soil Sample Classifications:

Soil Sample 1: pH = 3.1 (strongly acidic)
Soil Sample 2: pH = 7.2 (weakly basic)
Soil Sample 3: pH = 10.8 (strongly basic)

Item 5: Acid and Base Identification

The question requires identifying the acid and base in the reaction: $NH3 + HCl \rightarrow NH4^+ + Cl^-$ .
Acid: A substance that donates a proton (H+).
Base: A substance that accepts a proton (H+).
In this reaction, HCl donates a proton to $NH_3$ , so:
- HCl is the acid.
- $NH_3$ is the base.

Item 6: Heat Transfer and Heat Capacity

A 10 g metal sample at 100°C is transferred to 1000 g of water at 25°C, and the final temperature is 50°C.

Item 6a: Temperature Change

The metal experienced a larger temperature change because it went from 100°C to 50°C (change of 50°C), while the water went from 25°C to 50°C (change of 25°C).

Item 6b: Heat/Thermal Energy

The water gained more heat/thermal energy. This is because heat gained by water equals heat lost by the metal: $Q = mc\Delta T$ . The mass of the water is significantly larger (1000g vs 10g), and while its temperature change is smaller, the large mass compensates for it.

Item 6c: Heat Capacity

The metal has a smaller heat capacity than water. Heat capacity is the amount of heat required to raise the temperature of a substance by 1 degree Celsius. Since the metal's temperature changed more significantly with less mass, it indicates a lower heat capacity.

Item 7: Stoichiometry and Gas Laws

The question involves completing reactions to determine the amount of aluminum oxide ( $Al2O3$ ) formed.

Item 7a: Moles of Oxygen Gas

Using the ideal gas law: $PV = nRT$ , where:
- P = 1.2 atm
- V = 2 L
- R = 0.0821 L atm / (mol K)
- T = 30°C = 303 K
Solving for n:
$n = \frac{PV}{RT} = \frac{1.2 \, atm \times 2 \, L}{0.0821 \, L \cdot atm/(mol \cdot K) \times 303 \, K} = 0.096 \, mol$

Item 7b: Grams of Oxygen Gas

Moles of $O_2$ = 0.096 mol
Molar mass of $O_2$ = 32 g/mol
Mass of $O_2$ : $m = 0.096 \, mol \times 32 \, g/mol = 3.072 \, g$

Item 7c: Mass of $Al2O3$ Formed (Stoichiometry)

The balanced reaction for the formation of aluminum oxide from aluminum and oxygen is: $4Al + 3O2 \rightarrow 2Al2O_3$
Starting with 3.072 g of $O2$ , calculate the grams of $Al2O_3$ that can be formed:
- Moles of $O_2$ = 0.096 mol (calculated above)
- From the balanced equation, 3 moles of $O2$ produce 2 moles of $Al2O_3$ .
- Moles of $Al2O3$ : $n(Al2O3) = \frac{2}{3} \times 0.096 \, mol = 0.064 \, mol$
- Molar mass of $Al2O3$ = 101.96 g/mol
- Mass of $Al2O3$ : $m = 0.064 \, mol \times 101.96 \, g/mol = 6.525 \, g$

Item 8: Calorimetry and Heat Energy

The question involves using a bomb calorimeter to determine the heat energy in a burger sample.
Given: 10 g burger sample, 500 g water, initial temperature = 21°C, final temperature = 77°C.
The formula to determine heat energy (Q) is: $Q = mc\Delta T$
- m = mass of water = 500 g
- c = specific heat capacity of water = 4.184 J/(g·°C)
- $\Delta T$ = change in temperature = 77°C - 21°C = 56°C
$Q = 500 \, g \times 4.184 \, J/(g \cdot \, ^\circ C) \times 56 \, ^\circ C = 117152 \, J$

Item 9: Laws of Conservation

Law of Conservation of Mass: Matter cannot be created or destroyed.
Law of Conservation of Energy: Energy cannot be created or destroyed, only transformed.
Experimental Proof of Conservation of Mass: Perform a chemical reaction in a closed container and measure the mass before and after the reaction. The mass should remain the same, demonstrating that mass is conserved.

Item 10: Factors Affecting Reaction Rate

Crushing a Solid: Increases the surface area, leading to more collisions between reactant molecules, thus increasing the reaction rate.
Heating the Reactions: Increases the kinetic energy of the molecules, resulting in more frequent and energetic collisions, increasing the reaction rate.
Lowering Volume of Container: Increases the concentration of reactants, leading to more frequent collisions, thus increasing the reaction rate.
Adding More Reactant: Increases the concentration of reactants, leading to more frequent collisions, thus increasing the reaction rate.

Item 10: Exothermic Reactions and Bond Energies

For an exothermic reaction (energy is released), the energy required to break the bonds of the reactants is lesser than the energy released when forming the bonds of the products.

Item 11: Ideal Gas Law

Using the ideal gas law: $PV = nRT$ , where:
- P = 0.95 atm
- n = 12.1 moles
- R = 0.0821 L atm / (mol K)
- T = 29°C = 302 K
Solving for V:
$V = \frac{nRT}{P} = \frac{12.1 \, mol \times 0.0821 \, L \cdot atm/(mol \cdot K) \times 302 \, K}{0.95 \, atm} = 315.5 \, L$

Item 12: Ideal Gas Law - Solving for Moles

Using the ideal gas law: $PV = nRT$ , where:
- P = 0.95 atm
- V = 20 L
- R = 0.0821 L atm / (mol K)
- T = 29°C = 302 K
Solving for n:
$n = \frac{PV}{RT} = \frac{0.95 \, atm \times 20 \, L}{0.0821 \, L \cdot atm/(mol \cdot K) \times 302 \, K} = 0.766 \, mol \approx 0.77 \, mol$

Item 13: Kinetic Molecular Theory and Balloon Volume

As the temperature decreases, the balloon's volume will decrease.
Justification via Kinetic Molecular Theory (KMT):
- KMT states that gas particles are in constant, random motion. As temperature decreases, the average kinetic energy of the gas particles decreases.
- With less kinetic energy, the particles collide with the walls of the balloon with less force and frequency.
- The external atmospheric pressure remains relatively constant. Therefore, the balloon's volume decreases until the internal pressure matches the external pressure.

Item 14: Molarity Calculation

Given: 1.2 L solution of 0.8 M HCl.
Using the formula: $M = \frac{mol}{L}$
Solving for moles:
$mol = M \times L = 0.8 \, M \times 1.2 \, L = 0.96 \, mol$

Item 15: Periodic Trends (Nitrogen vs. Oxygen)

Ionization Energy: The energy required to remove an electron from an atom.
Atomic Radius: The size of an atom.
Effective Nuclear Charge: The net positive charge experienced by an electron in an atom.

Comparison between Nitrogen (N) and Oxygen (O):

Ionization Energy: Oxygen has a larger ionization energy than Nitrogen.
- Justification: Oxygen has a greater effective nuclear charge than Nitrogen because it has one more proton in its nucleus. This stronger positive charge attracts the electrons more strongly, making it harder to remove an electron.
Atomic Radius: Nitrogen has a larger atomic radius than Oxygen.
- Justification: Although Oxygen has more protons, the electrons are pulled closer to the nucleus due to the greater effective nuclear charge, resulting in a smaller atomic radius. Nitrogen's weaker effective nuclear charge results in a slightly larger atomic radius.

Item 16: Heat Capacity and Temperature Change

Equal amounts of energy are added to substances A and B with equal masses and initial temperatures. The final temperature of A is higher than B.

Item 16a: Heat Capacity

The heat capacity of substance A was less than the heat capacity of substance B.
Explanation: Heat capacity is the amount of heat required to raise the temperature of a substance. Substance A reached a higher final temperature with the same amount of energy, indicating it requires less heat to change its temperature.

Item 16b: Heat Absorbed

The total amount of heat absorbed by substance A was equal to the amount of energy absorbed by substance B.
Explanation: The question states that equal amounts of energy were added to both substances.

Item 16c: Temperature Change

The total temperature change of substance A was greater than the total temperature change of substance B.
Explanation: The final temperature of A was higher than that of B, and they started at the same initial temperature, so A had a greater temperature change.

Item 17: Intermolecular Forces (IMFs)

Comparing Ammonia (NH3) and Methane (CH4).

Item 17a: Intermolecular Forces Present

Ammonia (NH3): Hydrogen bonding, Dipole-dipole forces, London dispersion forces
- Hydrogen bonding occurs because hydrogen is bonded to a highly electronegative atom (nitrogen).
Methane (CH4): London dispersion forces only
- Methane is nonpolar due to its symmetrical tetrahedral shape, so it only exhibits London dispersion forces.

Item 17b: Strength of IMFs

Ammonia (NH3) has larger and stronger intermolecular forces overall.
Justification: Ammonia exhibits hydrogen bonding, which is a much stronger IMF than London dispersion forces (the only IMF present in methane). The higher boiling point of ammonia (-28°C) compared to methane (-161.6°C) indicates that more energy is required to overcome the IMFs in ammonia, hence they are stronger.

Item 18: Heat Energy Calculation from Graph

The question involves a graph of temperature versus time for an object being heated.

Item 18a: Change in Temperature

To calculate the change in temperature, subtract the initial temperature from the final temperature. Without the specific graph values, assume initial T = $Ti$ and final T = $Tf$ . $\Delta T = Tf - Ti$

Item 18b: Heat Energy Absorbed

Given: mass (m) = 1.0 g, specific heat capacity (Cp) = 2.5 J/(g·°C).
The heat energy absorbed (Q) is calculated using: $Q = mc\Delta T$
If we assume a temperature change of, for example, 20°C : $\Delta T = 20 \, ^\circ C$
$Q = 1.0 \, g \times 2.5 \, J/(g \cdot \, ^\circ C) \times 20 \, ^\circ C = 50 \, J$