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Chapter 1: Problem 16

Nitrogen gas was collected over water at \(25^{\circ} \mathrm{C}\) . If the vapor pressure of water at \(25^{\circ} \mathrm{C}\) is 23 \(\mathrm{mmH}\) g, and the total pressure in the container is measured at 781 \(\mathrm{mmH} \mathrm{g}\) , what is the partial pressure of the nitrogen gas? \(\begin{array}{ll}{\text { (A) }} & {46 \mathrm{mmH} \mathrm{g}} \\ {\text { (B) }} & {551 \mathrm{mmH} \mathrm{g}} \\ {\text { (C) }} & {735 \mathrm{mmH} \mathrm{g}} \\ {\text { (D) }} & {758 \mathrm{mmH} \mathrm{g}}\end{array}\)

Short Answer

Expert verified
The partial pressure of the nitrogen gas is \(758 mmHg\). So, the answer is (D) \(758 mmHg\).

Step by step solution

01

Understand the total pressure

The total pressure inside the container is a sum of the pressure of the nitrogen and the vapor pressure of the water. This is given as 781 mmHg.
02

Note the vapor pressure of water

We are told that the vapor pressure of water at the given temperature (25°C) is 23 mmHg. This is the pressure that the water exerts in the container.
03

Calculate the partial pressure of the nitrogen gas

The partial pressure of the nitrogen gas can now be found by subtracting the vapor pressure of the water from the total pressure. Mathematically, this can be expressed as: \( P(N_{2}) = P(total) - P(water) \) Substituting the given values: \( P(N_{2}) = 781 mmHg - 23 mmHg = 758 mmHg \)

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Most popular questions from this chapter

Molten \(\mathrm{AlCl}_{3}\) is electrolyzed with a constant current of 5.00 amperes over a period of 600.0 seconds. Which of the following expressions is equal to the maximum mass of Al(s) that plates out? (1 faraday = 96,500 coulombs) (A) \(\frac{(600)(5.00)}{(96,500)(3)(27.0)}\) grams (B) \(\frac{(600)(5.00)(3)(27.0)}{(96,500)}\) grams (C) \(\frac{(600)(5.00)(27.0)}{(96,500)(3)}\) grams (D) \(\frac{(96,500)(3)(27.0)}{(600)(5.00)}\) grams

$$2 \mathrm{NOCl} \rightarrow 2 \mathrm{NO}+\mathrm{Cl}_{2}$$ The reaction above takes place with all of the reactants and products in the gaseous phase. Which of the following is true of the relative rates of disappearance of the reactants and appearance of the products? (A) NO appears at twice the rate that NOCl disappears. (B) NO appears at the same rate that NOCl disappears. (C) NO appears at half the rate that NOCl disappears. (D) \(\mathrm{Cl}_{2}\) appears at the same rate that NOCl disappears.

$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The reaction above came to equilibrium at a temperature of \(100^{\circ} \mathrm{C} .\) At equilibrium the partial pressure due to NOBr was 4 atmospheres, the partial pressure due to NO was 4 atmospheres, and the partial pressure due to \(\mathrm{Br}_{2}\) was 2 atmospheres. What is the equilibrium constant, \(K_{p},\) for this reaction at \(100^{\circ} \mathrm{C}\) ? (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) 1 (D) 2

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) \(\frac{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}{2\left[\mathrm{SO}_{3}\right]}\)

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. As \(\Delta T\) increases, what happens to the equilibrium constant and why? (A) The equilibrium constant increases because more products are created. (B) The equilibrium constant increases because the rate of the forward reaction increases. (C) The equilibrium constant decreases because the equilibrium shifts to the left. (D) The value for the equilibrium constant is unaffected by temperature and will not change.
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