Exam 2 practice

Core Knowledge and Skills Breakdown

Stoichiometry and Mole Calculations
- Molar Mass Calculation
  - Summing the atomic masses of each element in a formula to obtain molar mass in $g\,mol^{-1}$ .
- Mole and Mass Conversions
  - Using molar mass to convert between grams and moles: $n = \frac{m}{M}$ .
  - Applying Avogadro's number ( $6.02 \times 10^{23}\,mol^{-1}$ ) to determine particle counts.
- Reaction Stoichiometry
  - Using balanced chemical equations to determine mole ratios and limiting reactants.
  - Calculating theoretical yield of products from a given mass of reactant.
- Equivalents
  - Calculating equivalents based on ionic charges: $\text{Equivalents} = \text{moles} \times |\text{charge}|$
Gas Behavior and Laws
- Gas Variables and Conversions
 - Pressure relationships: $1.00\,atm = 760\,torr = 760\,mmHg$
 - Absolute temperature conversion: $T(K) = T(^\circ\text{C}) + 273$
- Gas Laws
 - Boyle's Law (inverse relationship between pressure and volume): $P1V1 = P2V2$
 - Charles's Law (direct relationship between volume and absolute temperature): $\frac{V1}{T1} = \frac{V2}{T2}$
 - Combined Gas Law: $\frac{P1V1}{T1} = \frac{P2V2}{T2}$
 - Dalton's Law of Partial Pressures: $P{\text{total}} = P1 + P2 + P3 + \dots$
 - Molar volume of an ideal gas at STP: $22.4\,dm^3\,mol^{-1}$
Solutions and Concentrations
- Concentration Expressions
 - Molarity ( $mol\,dm^{-3}$ ): $M = \frac{\text{moles of solute}}{\text{volume of solution in } dm^3}$
 - Mass/volume percent: $\frac{\text{mass of solute in } g}{\text{volume of solution in } cm^3} \times 100\%$
 - Mass percent ( $\% m/m$ ): $\frac{\text{mass of solute}}{\text{total mass of solution}} \times 100\%$
- Dilution Calculations
 - Dilution formula: $M1V1 = M2V2$
- Solution Properties
 - Distinguishing between strong electrolytes (complete dissociation), weak electrolytes (partial dissociation), and nonelectrolytes.
 - Distinguishing homogeneous solutions, colloids, and suspensions.
 - Saturated versus unsaturated solutions based on solubility limits.
 - Predicting precipitate formation using water solubility rules.
Acids, Bases, and Chemical Equilibria
- pH and Hydronium Calculations
 - Hydronium calculation: $\text{pH} = -\log([H_3O^+])$
 - Ion concentration calculation: $[H_3O^+] = 10^{-\text{pH}}$
- Conjugate Pairs and Buffers
 - Conjugate acid-base pairs differ by exactly one $H^+$ proton.
 - Buffer systems require a weak acid and its conjugate base (or a weak base and its conjugate acid).
- Acid-Base Neutralization
 - Stoichiometric titration reactions: $\text{acid} + \text{base} \rightarrow \text{salt} + H2O{(l)}$
- Le Châtelier's Principle
 - Predicting shifts in equilibrium when adding or removing species, altering volume/pressure, or changing temperature.
Chemical Reaction Types, Energy, and Tonicity
- Reaction Classifications
 - Single replacement, double replacement, combination, and decomposition reactions.
 - Oxidation-reduction (redox): loss of electrons is oxidation; gain of electrons is reduction.
- Thermochemistry
 - Endothermic reactions absorb energy ( $\Delta H > 0$ ); exothermic reactions release energy ( $\Delta H < 0$ ).
- Tonicity and Osmosis
 - Hypertonic solutions cause red blood cells to lose water and undergo crenation.
 - Hypotonic solutions cause cell swelling and hemolysis; isotonic solutions maintain cell equilibrium.

Targeted Practice Test

What is the molar mass of magnesium hydroxide, $Mg(OH)_2$ ?
- A) $58.3\,g\,mol^{-1}$
- B) $41.3\,g\,mol^{-1}$
- C) $82.6\,g\,mol^{-1}$
- D) $74.1\,g\,mol^{-1}$
In the precipitation reaction $AgNO{3(aq)} + NaCl{(aq)} \rightarrow AgCl{(s)} + NaNO{3(aq)}$ , how many grams of $AgCl{(s)}$ ( $143.32\,g\,mol^{-1}$ ) form from $50.0\,g$ of $AgNO{3(aq)}$ ( $169.87\,g\,mol^{-1}$ ) with excess $NaCl_{(aq)}$ ?
- A) $21.1\,g$
- B) $42.2\,g$
- C) $84.4\,g$
- D) $58.9\,g$
A sample of neon gas occupies $250.0\,cm^3$ at $27\,^\circ\text{C}$ . What is the final volume of the gas if the temperature is raised to $127\,^\circ\text{C}$ at constant pressure?
- A) $187.5\,cm^3$
- B) $333.3\,cm^3$
- C) $500.0\,cm^3$
- D) $1173\,cm^3$
A sealed gas cylinder at $2.40\,atm$ occupies an initial volume of $3.00\,dm^3$ . If the container expands to a volume of $9.00\,dm^3$ at constant temperature, what is the final pressure?
- A) $0.800\,atm$
- B) $7.20\,atm$
- C) $1.20\,atm$
- D) $0.267\,atm$
What volume is occupied by $2.50\,mol$ of an ideal gas at STP?
- A) $8.96\,dm^3$
- B) $22.4\,dm^3$
- C) $56.0\,dm^3$
- D) $112\,dm^3$
What is the molarity of a solution prepared by dissolving $20.0\,g$ of $NaOH$ ( $40.00\,g\,mol^{-1}$ ) in enough water to prepare $0.250\,dm^3$ of solution?
- A) $0.500\,mol\,dm^{-3}$
- B) $1.00\,mol\,dm^{-3}$
- C) $2.00\,mol\,dm^{-3}$
- D) $4.00\,mol\,dm^{-3}$
What volume of a $3.00\,mol\,dm^{-3}$ $KCl$ stock solution is required to prepare $0.450\,dm^3$ of a $0.200\,mol\,dm^{-3}$ $KCl$ solution?
- A) $0.0300\,dm^3$
- B) $0.0675\,dm^3$
- C) $0.150\,dm^3$
- D) $0.0150\,dm^3$
A solution has a hydronium ion concentration of $[H_3O^+] = 1.0 \times 10^{-4}\,mol\,dm^{-3}$ . What is the pH of this solution?
- A) $10.0$
- B) $4.00$
- C) $1.00$
- D) $7.00$
Which of the following pairs is a conjugate acid-base pair?
- A) $HCl$ and $NaOH$
- B) $H2CO3$ and $HCO_3^-$
- C) $H_3O^+$ and $OH^-$
- D) $H2SO4$ and $SO_4^{2-}$
In the reaction $Zn{(s)} + Cu(NO3){2(aq)} \rightarrow Zn(NO3){2(aq)} + Cu{(s)}$ , which substance is oxidized?
- A) $Cu^{2+}$
- B) $NO_3^-$
- C) $Zn_{(s)}$
- D) $Cu_{(s)}$
When red blood cells are placed into a $5.0\%$ $NaCl$ solution, water exits the cells, causing them to shrink. This cellular process is known as:
- A) Hemolysis
- B) Dialysis
- C) Neutralization
- D) Crenation
For the endothermic equilibrium reaction $N2O{4(g)} \rightleftharpoons 2NO_{2(g)}$ , which adjustment will shift the equilibrium position toward the products?
- A) Increasing the temperature
- B) Decreasing the temperature
- C) Increasing the pressure
- D) Removing $N2O{4(g)}$

Practice Test Answer Key and Explanations

A) $58.3\,g\,mol^{-1}$
- Calculation: $24.31 + 2(16.00) + 2(1.008) = 58.33\,g\,mol^{-1}$
B) $42.2\,g$
- Calculation: $\frac{50.0\,g}{169.87\,g\,mol^{-1}} = 0.2943\,mol\,AgNO_3$
- Yield: $0.2943\,mol \times 143.32\,g\,mol^{-1} = 42.2\,g\,AgCl$
B) $333.3\,cm^3$
- Temperature conversion: $T1 = 27 + 273 = 300\,K$ ; $T2 = 127 + 273 = 400\,K$
- Charles's Law: $V2 = \frac{V1 T2}{T1} = \frac{250.0\,cm^3 \times 400\,K}{300\,K} = 333.3\,cm^3$
A) $0.800\,atm$
- Boyle's Law: $P2 = \frac{P1 V1}{V2} = \frac{2.40\,atm \times 3.00\,dm^3}{9.00\,dm^3} = 0.800\,atm$
C) $56.0\,dm^3$
- Calculation: $2.50\,mol \times 22.4\,dm^3\,mol^{-1} = 56.0\,dm^3$
C) $2.00\,mol\,dm^{-3}$
- Moles: $\frac{20.0\,g}{40.00\,g\,mol^{-1}} = 0.500\,mol$
- Molarity: $\frac{0.500\,mol}{0.250\,dm^3} = 2.00\,mol\,dm^{-3}$
A) $0.0300\,dm^3$
- Dilution formula: $V1 = \frac{M2 V2}{M1} = \frac{0.200\,mol\,dm^{-3} \times 0.450\,dm^3}{3.00\,mol\,dm^{-3}} = 0.0300\,dm^3$
B) $4.00$
- Calculation: $\text{pH} = -\log(1.0 \times 10^{-4}) = 4.00$
B) $H2CO3$ and $HCO_3^-$
- Explanation: A conjugate pair differs strictly by the transfer of one hydrogen ion ( $H^+$ ).
C) $Zn_{(s)}$
- Explanation: Zinc loses two electrons as its oxidation state increases from $0$ to $+2$ , meaning it is oxidized.
D) Crenation
- Explanation: A $5.0\%$ $NaCl$ solution is hypertonic relative to the cell's interior, drawing water out by osmosis and causing crenation.
A) Increasing the temperature
- Explanation: Because the reaction absorbs heat (endothermic), adding heat shifts the equilibrium position toward the forward endothermic direction to produce more products.