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Chapter 1: Problem 23

Use the following information to answer questions 22-24 10.0 g each of three different gases are present in three glass containers of identical volume, as shown below. The temperature of all three flasks is held constant at 298 K. Which of the gases would have the greatest density? (A) \(\mathrm{SO}_{2}\) (B) \(\mathrm{CH}_{4}\) (C) \(\mathrm{NCl}_{3}\) (D) All three gases would have the same density.

Short Answer

Expert verified
So, the gas with the greatest density is \( \mathrm{NCl}_{3} \).

Step by step solution

01

Determine the Molar Mass of Each Gas

Use the periodic table to find the atomic masses of the elements in each molecule. The molar mass of \( \mathrm{SO}_{2} \) is \( 32.06 (S) + 2 \cdot 16.00 (O) = 64.06 \) g/mol. The molar mass of \( \mathrm{CH}_{4} \) is \( 12.01 (C) + 4 \cdot 1.008 (H) = 16.04 \) g/mol. The molar mass of \( \mathrm{NCl}_{3} \) is \( 14.01 (N) + 3 \cdot 35.45 (Cl) = 120.36 \) g/mol.
02

Compare Molar Masses

The greater the molar mass, the greater the density of a gas. By comparing the molar masses calculated in Step 1, \( \mathrm{NCl}_{3} \) with the molar mass of 120.36 g/mol has the greatest molar mass, meaning it will have the greatest density.

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Most popular questions from this chapter

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ How many moles of electrons must be transferred to create 127 g of copper? (A) 1 mole of electrons (B) 2 moles of electrons (C) 3 moles of electrons (D) 4 moles of electrons

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) What is the expression for the equilibrium constant, \(K_{\mathrm{c}} ?\) (A) \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}\) (B) \(\frac{2\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}\) (C) \(\frac{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{SO}_{3}\right]^{2}}\) (D) \(\frac{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}{2\left[\mathrm{SO}_{3}\right]}\)

A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ As the reaction progresses, what will happen to the overall voltage of the cell? (A) It will increase as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (B) It will increase as \(\left[\mathrm{Cu}^{+}\right]\) increases. (C) It will decrease as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (D) The voltage will remain constant.
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