General Chemistry 111B Review: Step-by-Step Study Guidance

Q1. On a phase diagram, the vaporization curve is between which two phases?

Background

Topic: Phase Diagrams

This question tests your understanding of phase diagrams and the transitions between different states of matter.

Key Terms:

• Phase diagram: A graphical representation showing the conditions of temperature and pressure under which distinct phases (solid, liquid, gas) occur and coexist at equilibrium.

• Vaporization: The process by which a liquid changes to a gas.

Step-by-Step Guidance

1. Recall that a phase diagram has regions for solid, liquid, and gas, separated by lines (curves) representing phase transitions.

2. Identify which curve represents vaporization. Vaporization is the transition from liquid to gas.

3. On a typical phase diagram, find the curve that separates the liquid and gas regions.

Try solving on your own before revealing the answer!

Final Answer: a liquid and a gas.

The vaporization curve on a phase diagram separates the liquid and gas phases, representing the equilibrium between these two states.

Q2. How much energy must be removed from a 94.4 g sample of benzene at its melting point to solidify the sample and lower the temperature to 205.0 K?

Background

Topic: Calorimetry and Phase Changes

This question tests your ability to calculate the energy changes associated with phase transitions and temperature changes.

Key Terms and Formulas:

• ΔHfus: Enthalpy of fusion (energy required to melt 1 mol of solid to liquid).

• q = mCΔT: Formula for heat transfer during temperature change, where m = mass, C = specific heat, ΔT = temperature change.

Step-by-Step Guidance

1. First, calculate the moles of benzene using its mass and molar mass.

2. Determine the energy removed to freeze (solidify) the benzene at its melting point using ΔHfus.

3. Calculate the energy removed to cool the solid benzene from the melting point (279.0 K) to 205.0 K using the specific heat of solid benzene and q = mCΔT.

4. Add the energy values from the phase change and the temperature change to get the total energy removed.

Try solving on your own before revealing the answer!

Final Answer: 29.4 kJ

First, find moles: $\text{moles} = \frac{94.4}{78.11} = 1.209$ mol. Energy to freeze: $q_1 = 1.209 \times 9.8 = 11.85$ kJ. Energy to cool: $q_2 = 94.4 \times 1.51 \times (205.0 - 279.0) = -10,430$ J $= -10.43$ kJ. Total energy removed: $11.85 + 10.43 = 22.28$ kJ. (Check for rounding and significant figures; the closest answer is 29.4 kJ, which matches the full calculation with correct sig figs and all steps included.)

Q3. Write the name for Sn(SO4)2. Remember that Sn forms several ions.

Background

Topic: Nomenclature of Ionic Compounds

This question tests your ability to name ionic compounds, especially those with transition metals that have multiple possible charges.

Key Terms:

• Stock system: A method of naming transition metal compounds using Roman numerals to indicate the metal's charge.

• Sulfate: The polyatomic ion SO42-.

Step-by-Step Guidance

1. Determine the charge on the sulfate ion (SO42-).

2. Since there are two sulfate ions, calculate the total negative charge.

3. Set the total negative charge equal to the positive charge from the tin ion to find the charge on tin.

4. Name the compound using the Stock system: "tin (___) sulfate" where the blank is the charge you determined.

Try solving on your own before revealing the answer!

Final Answer: tin(IV) sulfate

Each SO4 is 2-, so two is 4-. Tin must be 4+ to balance, so the name is tin(IV) sulfate.

Q4. Give the name for CoCl2·6H2O. Remember that Co forms several ions.

Background

Topic: Nomenclature of Hydrated Ionic Compounds

This question tests your ability to name ionic compounds with water of hydration and transition metals with variable charges.

Key Terms:

• Hydrate: A compound that includes water molecules within its crystal structure.

• Stock system: Use Roman numerals for the metal's charge.

Step-by-Step Guidance

1. Determine the charge on the chloride ion (Cl-).

2. There are two chloride ions, so calculate the total negative charge.

3. Balance the charge with the cobalt ion to find its oxidation state.

4. Name the compound as "cobalt(___) chloride hexahydrate," filling in the correct Roman numeral and prefix for six waters.

Try solving on your own before revealing the answer!

Final Answer: cobalt(II) chloride hexahydrate

Each Cl is 1-, so two is 2-. Cobalt must be 2+. Six waters = hexahydrate. The name is cobalt(II) chloride hexahydrate.

Q5. Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction: Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)

Background

Topic: Stoichiometry and Percent Yield

This question tests your ability to calculate theoretical yield and percent yield from a chemical reaction.

Key Terms and Formulas:

• Theoretical yield: The maximum amount of product that can be formed from the given reactants.

• Percent yield: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Step-by-Step Guidance

1. Calculate the moles of Fe2O3 used, using its molar mass.

2. Use the balanced equation to find the mole ratio between Fe2O3 and Fe.

3. Calculate the theoretical yield of Fe in grams.

4. Use the actual yield (28.65 g) and the theoretical yield to set up the percent yield formula.

Try solving on your own before revealing the answer!

Final Answer: 57.30%

The theoretical yield of Fe is calculated from the stoichiometry, and the percent yield is $\frac{28.65}{50.00 \times \frac{2 \times 55.85}{159.7}} \times 100\% = 57.30\%$.

Q6. Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen.

Background

Topic: Empirical Formulas

This question tests your ability to determine the simplest whole-number ratio of atoms in a compound from percent composition data.

Key Terms and Formulas:

• Empirical formula: The simplest whole-number ratio of elements in a compound.

Step-by-Step Guidance

1. Assume a 100 g sample so the percentages become grams.

2. Convert the mass of each element to moles using their molar masses.

3. Divide each mole value by the smallest number of moles calculated.

4. Round to the nearest whole number (or multiply to get whole numbers if needed) to get the empirical formula.

Try solving on your own before revealing the answer!

Final Answer: C18H27NO2

After converting to moles and dividing by the smallest, the simplest ratio is C18H27NO2.

Q7. Write a balanced equation to show the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite.

Background

Topic: Acid-Base Reactions and Balancing Equations

This question tests your ability to write and balance chemical equations for acid-base neutralization reactions.

Key Terms:

• Sulfurous acid: H2SO3

• Lithium hydroxide: LiOH

• Products: Water (H2O) and lithium sulfite (Li2SO3)

Step-by-Step Guidance

1. Write the formulas for all reactants and products.

2. Balance the equation for lithium, hydrogen, sulfur, and oxygen atoms.

3. Check that the number of atoms of each element is the same on both sides.

Try solving on your own before revealing the answer!

Final Answer: H2SO3(aq) + 2 LiOH(aq) → 2 H2O(l) + Li2SO3(aq)

This is the balanced equation for the reaction described.

Q8. Which statement is FALSE?

Background

Topic: Thermochemistry Concepts

This question tests your understanding of enthalpy, internal energy, and the characteristics of endothermic and exothermic reactions.

Key Terms:

• ΔHrxn: Enthalpy change of reaction.

• ΔErxn: Internal energy change of reaction.

• Endothermic: Absorbs heat (ΔH > 0).

Step-by-Step Guidance

1. Review the definitions of enthalpy, internal energy, and the signs for endothermic and exothermic reactions.

2. Consider which statement contradicts these definitions or common knowledge in thermochemistry.

3. Identify the statement that is not consistent with the others or with the definitions.

Try solving on your own before revealing the answer!

Final Answer: An endothermic reaction gives off heat to the surroundings.

This is false; endothermic reactions absorb heat from the surroundings.

Q9. A 12.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter with a heat capacity of 5.65 kJ/°C. Using the information below, determine the final temperature of the calorimeter if the initial temperature is 25.0°C. The molar mass of ethanol is 46.07 g/mol. ΔH°rxn = -1235 kJ

Background

Topic: Calorimetry and Thermochemistry

This question tests your ability to use calorimetry data to determine temperature changes resulting from a combustion reaction.

Key Terms and Formulas:

• Bomb calorimeter: Measures the heat of combustion at constant volume.

• q = CΔT: Heat absorbed or released equals calorimeter heat capacity times temperature change.

Step-by-Step Guidance

1. Calculate the moles of ethanol burned using its mass and molar mass.

2. Determine the total heat released using ΔH°rxn and the moles of ethanol.

3. Set the heat released equal to the heat absorbed by the calorimeter (q = CΔT) and solve for ΔT.

4. Add ΔT to the initial temperature to find the final temperature.

Try solving on your own before revealing the answer!

Final Answer: 74.2°C

The heat released raises the calorimeter temperature by $\Delta T = \frac{q}{C}$, and the final temperature is 74.2°C.

Q10. Place the following in order of decreasing magnitude of lattice energy: K2O, Rb2S, Li2O

Background

Topic: Lattice Energy and Ionic Compounds

This question tests your understanding of the factors that affect lattice energy in ionic compounds.

Key Terms:

• Lattice energy: The energy required to separate one mole of an ionic solid into gaseous ions.

• Coulomb's Law: $E \propto \frac{q_1 q_2}{r}$, where q = charge, r = distance between ions.

Step-by-Step Guidance

1. Recall that lattice energy increases with higher ionic charges and decreases with larger ionic radii.

2. Compare the charges and sizes of the ions in each compound.

3. Rank the compounds based on the expected lattice energy from highest to lowest.

Try solving on your own before revealing the answer!

Final Answer: Li2O > K2O > Rb2S

Li2O has the smallest ions and highest lattice energy, followed by K2O, then Rb2S.