General Chemistry Practice Exam 3 – Step-by-Step Study Guidance

Q1. Place the following elements in order of decreasing atomic radius: Xe, Cs, Ne.

Background

Topic: Periodic Trends – Atomic Radius

This question tests your understanding of how atomic radius changes across periods and down groups in the periodic table.

Key Terms and Concepts:

• Atomic radius: The distance from the nucleus to the outermost electron shell of an atom.

• Periodic trend: Atomic radius increases down a group and decreases across a period (left to right).

Step-by-Step Guidance

1. Locate Xe (xenon), Cs (cesium), and Ne (neon) on the periodic table. Identify their periods and groups.

2. Recall that atomic radius increases as you move down a group and decreases as you move from left to right across a period.

3. Compare the positions: Cs is in group 1, period 6; Xe is in group 18, period 5; Ne is in group 18, period 2.

4. Determine which element should have the largest atomic radius based on its position (furthest down and furthest left).

Try solving on your own before revealing the answer!

Final Answer: Cs > Xe > Ne

Cesium is the largest because it is furthest down and to the left; neon is the smallest as it is in the top right.

Q2. Calculate the wavelength of light associated with the transition from n = 1 to n = 3 in the hydrogen atom.

Background

Topic: Atomic Structure – Hydrogen Emission Spectrum

This question tests your ability to use the Bohr model to calculate the wavelength of light emitted or absorbed during an electronic transition in hydrogen.

Key Formulas:

• $\Delta E = 2.18 \times 10^{-18} \text{ J} \left[ \left( \frac{1}{n_{\text{final}}^2} \right) - \left( \frac{1}{n_{\text{initial}}^2} \right) \right]$

• $E = h\nu$

• $c = \lambda \nu$

• Where $h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}$, $c = 3.00 \times 10^8 \text{ m/s}$

Step-by-Step Guidance

1. Identify $n_{\text{initial}} = 1$ and $n_{\text{final}} = 3$ for the transition.

2. Plug these values into the $\Delta E$ formula to find the energy change for the transition.

3. Remember: If the electron is moving from $n=1$ to $n=3$, energy is absorbed (positive $\Delta E$).

4. Once you have $\Delta E$, use $E = h\nu$ to solve for frequency $\nu$.

5. Then use $c = \lambda \nu$ to solve for wavelength $\lambda$.

Try solving on your own before revealing the answer!

Final Answer: 103 nm

The calculated energy corresponds to a wavelength of 103 nm for this transition.

Q3. Use the provided standard enthalpies of formation to calculate $\Delta H^\circ_{\text{rxn}}$ for: SO$_2$Cl$_2$(g) + 2 H$_2$O(l) → 2 HCl(g) + H$_2$SO$_4$(l)

Background

Topic: Thermochemistry – Enthalpy of Reaction

This question tests your ability to use standard enthalpies of formation to calculate the enthalpy change for a reaction.

Key Formula:

• $\Delta H^\circ_{\text{rxn}} = \sum n_p \Delta H^\circ_f(\text{products}) - \sum n_r \Delta H^\circ_f(\text{reactants})$

Step-by-Step Guidance

1. List the $\Delta H^\circ_f$ values for each compound and the number of moles from the balanced equation.

2. Multiply each $\Delta H^\circ_f$ by its coefficient in the equation.

3. Add up the total $\Delta H^\circ_f$ for products and for reactants separately.

4. Subtract the sum for reactants from the sum for products to find $\Delta H^\circ_{\text{rxn}}$.

Try solving on your own before revealing the answer!

Final Answer: –62 kJ

Using the formula and plugging in the values gives –62 kJ for the reaction.

Q4. Place the following in order of decreasing metallic character: P, As, K.

Background

Topic: Periodic Trends – Metallic Character

This question tests your understanding of how metallic character changes across the periodic table.

Key Terms:

• Metallic character: The tendency of an element to lose electrons and form positive ions (cations).

• Metallic character increases down a group and decreases across a period (left to right).

Step-by-Step Guidance

1. Locate P (phosphorus), As (arsenic), and K (potassium) on the periodic table.

2. Recall that metallic character increases as you move down a group and to the left across a period.

3. Compare the positions: K is an alkali metal, As and P are nonmetals/metalloids.

4. Arrange the elements from most metallic to least metallic based on their positions.

Try solving on your own before revealing the answer!

Final Answer: K > As > P

Potassium is the most metallic, followed by arsenic, then phosphorus.

Q5. A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO$_3$ in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine $\Delta H^\circ_{\text{rxn}}$ (in kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) no heat is lost to the calorimeter or surroundings, and 2) the density and heat capacity of the resulting solution are the same as water.

Background

Topic: Thermochemistry – Calorimetry and Enthalpy of Neutralization

This question tests your ability to use calorimetry data to calculate the enthalpy change per mole for a reaction.

Key Formulas:

• $q = mC_s\Delta T$

• $\Delta H_{\text{rxn}} = \frac{q}{\text{moles of NaOH}}$

• Assume $C_s$ (specific heat) of water = 4.18 J/g·°C, density = 1.00 g/mL

Step-by-Step Guidance

1. Calculate the total mass of the solution (sum of the two volumes, using density).

2. Calculate $q$ (heat absorbed or released) using $q = mC_s\Delta T$.

3. Determine the number of moles of NaOH used in the reaction.

4. Divide $q$ by the number of moles of NaOH to get $\Delta H_{\text{rxn}}$ in J/mol, then convert to kJ/mol.

Try solving on your own before revealing the answer!

Final Answer: –55.7 kJ/mol NaOH

The enthalpy change per mole of NaOH is –55.7 kJ/mol, indicating an exothermic reaction.

Q6. Calculate the amount of heat (in kJ) required to raise the temperature of an 88.0-g sample of ethanol from 298.0 K to 405.0 K. The specific heat capacity of ethanol is 2.42 J/g°C.

Background

Topic: Thermochemistry – Heat Calculations

This question tests your ability to use the specific heat formula to calculate the heat required for a temperature change.

Key Formula:

• $q = mC_s\Delta T$

• Where $m$ = mass (g), $C_s$ = specific heat (J/g·°C), $\Delta T$ = change in temperature (°C)

Step-by-Step Guidance

1. Calculate $\Delta T$ by subtracting the initial temperature from the final temperature. (Remember to convert K to °C if needed.)

2. Plug the values for $m$, $C_s$, and $\Delta T$ into the formula $q = mC_s\Delta T$.

3. Calculate $q$ in joules, then convert to kilojoules by dividing by 1000.

Try solving on your own before revealing the answer!

Final Answer: 22.8 kJ

The heat required is 22.8 kJ for the given temperature change.

Q7. Identify the color of a flame test for lithium.

Background

Topic: Atomic Structure – Flame Tests

This question tests your knowledge of the characteristic colors produced by different elements in a flame test.

Key Concept:

• Each element emits a characteristic color when heated in a flame due to electronic transitions.

Step-by-Step Guidance

1. Recall the typical flame test colors for common alkali metals (Li, Na, K, etc.).

2. Associate lithium with its characteristic flame color.

Try solving on your own before revealing the answer!

Final Answer: Red

Lithium produces a red flame in a flame test.

Q8. Choose the ground state electron configuration for Ti$^{2+}$.

Background

Topic: Electronic Structure – Electron Configurations of Ions

This question tests your ability to write the electron configuration for a transition metal ion.

Key Concepts:

• Write the configuration for the neutral atom first, then remove electrons according to ion charge (from 4s before 3d for transition metals).

Step-by-Step Guidance

1. Write the electron configuration for neutral Ti (atomic number 22).

2. Remove two electrons to account for the 2+ charge, starting with the 4s orbital.

3. Write the resulting configuration in noble gas notation.

Try solving on your own before revealing the answer!

Final Answer: [Ar]3d$^2$

For Ti$^{2+}$, the two electrons are removed from the 4s orbital, leaving [Ar]3d$^2$.