General Chemistry Exam 1 Study Guidance

Q1. The mass of a diamond is usually expressed in carats, where 1 carat = 0.200 g. The density of diamond is 3.51 g/cm3. What is the volume of a 1.0 carat diamond?

Background

Topic: Density, Unit Conversion

This question tests your ability to use the density formula to relate mass and volume, and to convert between units as needed.

Key Terms and Formulas

• Density ($d$): The mass per unit volume of a substance.

• Formula: $d = \frac{m}{V}$

• Where: $d$ = density (g/cm3), $m$ = mass (g), $V$ = volume (cm3)

Step-by-Step Guidance

1. Convert the mass of the diamond from carats to grams using the given conversion factor.

2. Write the density formula and rearrange it to solve for volume: $V = \frac{m}{d}$

3. Plug in the mass (in grams) and the density (in g/cm3) into the formula.

4. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 0.0570 cm3

$V = \frac{0.200\ \text{g}}{3.51\ \text{g/cm}^3} = 0.0570\ \text{cm}^3$

The mass was converted to grams, and the density formula was used to find the volume.

Q2. The coolant in an automobile radiator freezes at -39°C and boils at 110°C. What is the freezing point in Fahrenheit?

Background

Topic: Temperature Conversion

This question tests your ability to convert temperatures between Celsius and Fahrenheit scales.

Key Terms and Formulas

• Temperature conversion formula: $T_{\text{F}} = \frac{9}{5}T_{\text{C}} + 32$

• Where: $T_{\text{F}}$ = temperature in Fahrenheit, $T_{\text{C}}$ = temperature in Celsius

Step-by-Step Guidance

1. Identify the temperature in Celsius to be converted: $T_{\text{C}} = -39$°C.

2. Write the conversion formula for Celsius to Fahrenheit.

3. Plug the Celsius value into the formula: $T_{\text{F}} = \frac{9}{5} \times (-39) + 32$

4. Set up the calculation for the multiplication and addition, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: -38 °F

$T_{\text{F}} = \frac{9}{5} \times (-39) + 32 = -38$°F$

The formula was applied correctly to convert from Celsius to Fahrenheit.

Q3. The symbol for a cation with a mass number of 56, an atomic number of 30, and a charge of 2+ is:

Background

Topic: Atomic Structure, Isotope Notation

This question tests your understanding of how to write the symbol for an ion, including mass number, atomic number, and charge.

Key Terms and Formulas

• Atomic number ($Z$): Number of protons in the nucleus.

• Mass number ($A$): Total number of protons and neutrons.

• Isotope notation: $^{A}_{Z}\text{Element}^{\text{charge}}$

Step-by-Step Guidance

1. Identify the element with atomic number 30 using the periodic table.

2. Write the isotope notation with mass number (56) as the superscript and atomic number (30) as the subscript.

3. Add the charge (2+) as a superscript to the right of the symbol.

4. Compare your answer to the choices provided.

Try solving on your own before revealing the answer!

Final Answer: $^{56}_{30}\text{Zn}^{2+}$

The element with atomic number 30 is zinc (Zn), so the correct notation is $^{56}_{30}\text{Zn}^{2+}$.

Q4. How many moles of Au are there in 196.97 kg of Au?

Background

Topic: Moles, Molar Mass, Unit Conversion

This question tests your ability to convert mass to moles using the molar mass of gold (Au).

Key Terms and Formulas

• Mole: The amount of substance containing Avogadro's number of particles.

• Molar mass: The mass of one mole of a substance (g/mol).

• Formula: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

Step-by-Step Guidance

1. Convert the mass from kilograms to grams.

2. Find the molar mass of gold (Au) from the periodic table.

3. Plug the mass (in grams) and the molar mass into the formula for moles.

4. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 1000.0 mol

$\text{moles} = \frac{196970\ \text{g}}{196.97\ \text{g/mol}} = 1000.0\ \text{mol}$

The mass was converted to grams and divided by the molar mass of gold.

Q5. What percentage of CO2 is oxygen by mass?

Background

Topic: Percent Composition by Mass

This question tests your ability to calculate the percent by mass of an element in a compound.

Key Terms and Formulas

• Molar mass: The sum of the atomic masses of all atoms in a molecule.

• Percent by mass: $\%\ \text{by mass} = \frac{\text{mass of element in 1 mol compound}}{\text{molar mass of compound}} \times 100\%$

Step-by-Step Guidance

1. Calculate the molar mass of CO2 by adding the atomic masses of C and O (remember there are two O atoms).

2. Calculate the total mass of oxygen in one mole of CO2.

3. Divide the mass of oxygen by the molar mass of CO2 and multiply by 100% to get the percent by mass.

4. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 72.71 %

$\%\ \text{O} = \frac{2 \times 16.00}{44.01} \times 100\% = 72.71\%$

Oxygen's mass in CO2 is 32.00 g per mole, and the total molar mass is 44.01 g/mol.

Q6. A scientist was synthesizing ammonia with an expected yield of 97.8 kg. The amount produced, experimentally, was only 90.9 kg. What was his percent yield for this synthesis?

Background

Topic: Percent Yield

This question tests your ability to calculate percent yield from actual and theoretical yields.

Key Terms and Formulas

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

• Actual yield: The amount of product actually obtained from a reaction.

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

Step-by-Step Guidance

1. Identify the actual yield (90.9 kg) and the theoretical yield (97.8 kg).

2. Write the percent yield formula.

3. Plug the values into the formula: $\%\ \text{yield} = \frac{90.9}{97.8} \times 100\%$

4. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 92.9 %

$\%\ \text{yield} = \frac{90.9}{97.8} \times 100\% = 92.9\%$

The actual yield was divided by the theoretical yield and multiplied by 100%.

Q7. A recipe for 1 cup of hollandaise sauce calls for ½ cup of butter, ¼ cup of hot water, 4 egg yolks and the juice of a lemon. How many cups of this sauce can be made from 1 pound (2 cups) of butter, 2 dozen eggs, 12 lemons and an unlimited supply of hot water?

Background

Topic: Limiting Reactant (Stoichiometry in Cooking)

This question tests your ability to identify the limiting ingredient in a recipe, similar to limiting reactant problems in chemistry.

Key Terms and Formulas

• Limiting reactant: The ingredient that runs out first and limits the amount of product that can be made.

• Stoichiometry: The calculation of reactants and products in chemical reactions (or recipes).

Step-by-Step Guidance

1. List the amount of each ingredient available and the amount required per batch (per cup of sauce).

2. Calculate how many batches (cups) can be made from each ingredient by dividing the available amount by the required amount.

3. Identify the smallest number of batches possible; this is the limiting ingredient.

4. Set up the calculations for each ingredient, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 4 cups

Butter is the limiting ingredient: $2\ \text{cups} \div 0.5\ \text{cup per batch} = 4$ batches.

Eggs and lemons are in excess for 4 batches.

Q8. The concentration of Pb2+ in a stock solution is 1.000 g/L. What volume of this solution should be diluted to 500.0 mL to produce a solution in which the Pb2+ concentration is 100.0 mg/L?

Background

Topic: Dilution Calculations

This question tests your ability to use the dilution equation to determine the volume of stock solution needed to prepare a diluted solution.

Key Terms and Formulas

• Dilution equation: $C_1V_1 = C_2V_2$

• Where: $C_1$ = initial concentration, $V_1$ = volume of stock solution, $C_2$ = final concentration, $V_2$ = final volume

Step-by-Step Guidance

1. Convert all concentrations to the same units (e.g., mg/L or g/L).

2. Write the dilution equation and identify each variable.

3. Plug in the known values: $C_1 = 1.000$ g/L, $C_2 = 0.1000$ g/L (100.0 mg/L), $V_2 = 500.0$ mL (convert to L if needed).

4. Rearrange the equation to solve for $V_1$ (the volume of stock solution needed).

5. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 50.00 mL

$V_1 = \frac{C_2V_2}{C_1} = \frac{0.1000\ \text{g/L} \times 0.5000\ \text{L}}{1.000\ \text{g/L}} = 0.05000\ \text{L} = 50.00\ \text{mL}$

The dilution equation was used to find the required volume of stock solution.

Q9. When aqueous solutions of sodium chloride and silver nitrate are mixed, what happens to the solution?

Background

Topic: Precipitation Reactions, Solubility Rules

This question tests your understanding of double displacement reactions and the formation of precipitates.

Key Terms and Formulas

• Precipitate: An insoluble solid formed from a reaction in solution.

• Solubility rules: Guidelines for predicting whether an ionic compound will dissolve in water.

Step-by-Step Guidance

1. Write the balanced chemical equation for the reaction between sodium chloride and silver nitrate.

2. Use solubility rules to determine which products are soluble and which are insoluble.

3. Identify if a precipitate forms and, if so, which compound it is.

4. Match your findings to the answer choices.

Try solving on your own before revealing the answer!

Final Answer: It becomes cloudy due to the formation of solid silver chloride.

Silver chloride (AgCl) is insoluble and forms a white precipitate.

Q10. How many grams of sodium hydroxide are needed to create 100.00 mL of a 5.00 M solution?

Background

Topic: Solution Concentration (Molarity)

This question tests your ability to calculate the mass of solute needed to prepare a solution of a given molarity and volume.

Key Terms and Formulas

• Molarity ($M$): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Mass of solute: $\text{mass} = \text{moles} \times \text{molar mass}$

Step-by-Step Guidance

1. Convert the volume from mL to L.

2. Calculate the number of moles of NaOH needed using the molarity and volume.

3. Find the molar mass of NaOH.

4. Multiply the number of moles by the molar mass to get the mass in grams.

5. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 20.0 g

$\text{moles} = 5.00\ \text{mol/L} \times 0.10000\ \text{L} = 0.500\ \text{mol}$

$\text{mass} = 0.500\ \text{mol} \times 40.0\ \text{g/mol} = 20.0\ \text{g}$

Q11. After an early drive, the Curiosity rover on Mars stopped 4.0 ft away from an interesting football-shaped rock. If the rover moves at a speed of 200.0 m/sol, where 1 sol = 1 Martian day = 24.65 h, how many minutes would it take the rover to move to be 2.0 ft from the rock? (1 ft = 30.5 cm)

Background

Topic: Unit Conversion, Speed, and Time Calculations

This question tests your ability to convert between units and use the relationship between distance, speed, and time.

Key Terms and Formulas

• Speed: $\text{speed} = \frac{\text{distance}}{\text{time}}$

• Time: $\text{time} = \frac{\text{distance}}{\text{speed}}$

• Unit conversions: 1 ft = 30.5 cm, 1 m = 100 cm, 1 sol = 24.65 h, 1 h = 60 min

Step-by-Step Guidance

1. Calculate the distance the rover needs to travel: $4.0\ \text{ft} - 2.0\ \text{ft} = 2.0\ \text{ft}$

2. Convert the distance from feet to centimeters, then to meters.

3. Write the formula for time: $\text{time} = \frac{\text{distance}}{\text{speed}}$

4. Plug in the distance (in meters) and the speed (in m/sol) to find the time in sols.

5. Convert the time from sols to hours, then to minutes.

6. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: 8.99 min

The distance was converted to meters, divided by the speed to get time in sols, then converted to minutes.

Q12. Write balanced chemical reaction for each of the following:

Background

Topic: Writing and Balancing Chemical Equations

This question tests your ability to write and balance chemical equations for different types of reactions.

Key Terms and Formulas

• Reactants: Substances that undergo change in a chemical reaction.

• Products: Substances formed as a result of a chemical reaction.

• Balancing equations: Ensuring the same number of each type of atom on both sides of the equation.

Step-by-Step Guidance

1. For each reaction, write the correct chemical formulas for all reactants and products.

2. Arrange the reactants and products in a chemical equation.

3. Balance the equation by adjusting coefficients to ensure the same number of each atom on both sides.

4. Check your work for each equation.

Try writing and balancing each equation before revealing the answer!

Final Answers:

a. $2\text{Li}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{LiOH}(aq) + \text{H}_2(g)$

b. $\text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq)$

c. $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)$

Q13. Limonene is a hydrocarbon that contributes to the odor of citrus fruits, including lemons. Combustion of 0.671 g of limonene yielded 2.168 g of CO2 and 0.710 g of H2O. What is the empirical formula of limonene? If the molar mass of limonene is 136.24 g/mol; what is the molecular formula of limonene?

Background

Topic: Empirical and Molecular Formulas from Combustion Analysis

This question tests your ability to determine the empirical and molecular formulas of a compound from combustion data.

Key Terms and Formulas

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

• Molecular formula: The actual number of atoms of each element in a molecule.

• Combustion analysis: Used to determine the amounts of C and H in a hydrocarbon.

Step-by-Step Guidance

1. Calculate the moles of C from the mass of CO2 produced.

2. Calculate the moles of H from the mass of H2O produced.

3. Determine the mass of C and H in the original sample.

4. Subtract the mass of C and H from the total mass of limonene to find the mass of O (if present).

5. Convert the masses of each element to moles.

6. Find the simplest whole-number ratio of moles to determine the empirical formula.

7. Use the molar mass to determine the molecular formula.

8. Set up the calculations, but do not compute the final values yet.

Try working through the steps before revealing the answer!

Final Answer: Empirical formula C10H16; Molecular formula C10H16

The combustion data shows the ratio of C and H, and the molar mass confirms the molecular formula matches the empirical formula.

Q14. Acid rain in the form of aqueous H2SO4 slowly erodes marble statues. If we add 30.75 mL of 0.0100 M H2SO4 to 20.00 mg of CaCO3, will all of the CaCO3 react? Write a balanced chemical reaction for this process. You must show calculations to support your conclusion of which reactant is limiting to receive full credit.

Background

Topic: Limiting Reactant, Stoichiometry, Chemical Reactions

This question tests your ability to write a balanced chemical equation, perform stoichiometric calculations, and determine the limiting reactant.

Key Terms and Formulas

• Balanced chemical equation: Shows the reactants and products with correct stoichiometry.

• Limiting reactant: The reactant that is completely consumed first.

• Stoichiometry: Calculating the amounts of reactants and products in a chemical reaction.

Step-by-Step Guidance

1. Write the balanced chemical equation for the reaction between H2SO4 and CaCO3.

2. Calculate the moles of H2SO4 using its volume and molarity.

3. Calculate the moles of CaCO3 using its mass and molar mass.

4. Use the stoichiometry of the reaction to determine which reactant is limiting.

5. Set up the calculations, but do not compute the final values yet.

Try working through the steps before revealing the answer!

Final Answer: Not all CaCO3 will react; H2SO4 is the limiting reactant.

The stoichiometric calculations show that there is not enough H2SO4 to react with all the CaCO3 present.

Bonus: What are the seven diatomic elements? (Must get all 7.)

Background

Topic: Elements, Diatomic Molecules

This question tests your knowledge of the elements that naturally exist as diatomic molecules.

Key Terms and Formulas

• Diatomic element: An element that exists as a molecule composed of two atoms.

Step-by-Step Guidance

1. Recall the mnemonic "Have No Fear Of Ice Cold Beer" to remember the diatomic elements.

2. List the elements that exist as diatomic molecules in their natural state.

3. Write their chemical formulas (e.g., H2, N2, etc.).

Try listing all seven before revealing the answer!

Final Answer: H2, N2, O2, F2, Cl2, Br2, I2

The seven diatomic elements are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine.