/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 (a) Calculate the density of \(\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 49

(a) Calculate the density of \(\mathrm{NO}_{2}\) gas at \(0.970 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a gas if \(2.50 \mathrm{~g}\) occupies \(0.875 \mathrm{~L}\) at 685 torr and \(35^{\circ} \mathrm{C}\).

Short Answer

Expert verified
\( a) \) The density of NO鈧� gas at 0.970 atm and 35掳C is 1.77 g/L. \( b) \) The molar mass of the unknown gas is 73.1 g/mol.

Step by step solution

01

Part A: Calculate the moles of NO鈧�

\( P = 0.970 \ atm \\ T = 35 + 273.15 = 308.15 \ K \\ R = 0.0821 \frac{L \cdot atm}{mol \cdot K} \) Now we can use the molar mass of NO2: M(NO2) = 14.01 (for N) + 2 * 16.00 (for two O) = 46.01 g/mol. We can find the density using the formula: \[ d = \frac{MP}{RT} \] Where d is the density, M is the molar mass, P is the pressure, R is the ideal gas constant, and T is the temperature in Kelvin.
02

Calculate the density

Substitute the known values into the formula: \[ d = \frac{46.01 \frac{g}{mol} \cdot 0.970 \ atm}{0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 308.15 \ K} \] Now, calculate the density: \[ d = 1.77 \frac{g}{L} \] Thus, the density of NO鈧� gas at 0.970 atm and 35掳C is 1.77 g/L.
03

Part B: Calculate the moles of the unknown gas

For part (b), we are given the mass (m), volume (V), pressure (P), and temperature (T) of the unknown gas: \( m = 2.50 \ g \\ V = 0.875 \ L \\ P = 685 \ torr = 685 \frac{760 \ atm}{1 \ torr} = 0.901 \ atm \\ T = 35 + 273.15 = 308.15 \ K \\ R = 0.0821 \frac{L \cdot atm}{mol \cdot K} \) Use the Ideal Gas Law equation: \[ PV = nRT \] Solve for n: \[ n = \frac{PV}{RT} \]
04

Substitute the values

Substitute the given values into the formula: \[ n = \frac{(0.901 \ atm)(0.875 \ L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K})(308.15 \ K)} \] Calculate the number of moles: \[ n = 0.0342 \ mol \]
05

Calculate the molar mass

Now, we can use the mass and number of moles to find the molar mass (M) of the gas: \[ M = \frac{m}{n} \] Substitute the known values: \[ M = \frac{2.50 \ g}{0.0342 \ mol} \] Determine the molar mass: \[ M = 73.1 \frac{g}{mol} \] Hence, the molar mass of the unknown gas is 73.1 g/mol.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}\), \(15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2}\), and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(0.980 \mathrm{~atm}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C}\), calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2}\) ? (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2.)

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2}\) ? (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{NaClO}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ClO}_{2}(\mathrm{~g})+2 \mathrm{NaCl}(\mathrm{s}) $$ If you allow \(10.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(1.50 \mathrm{~atm}\) at \(21{ }^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

Perform the following conversions: (a) \(0.850\) atm to torr, (b) 785 torr to kilopascals, (c) \(655 \mathrm{~mm} \mathrm{Hg}\) to atmospheres, (d) \(1.323 \times 10^{5}\) Pa to atmospheres, (e) \(2.50\) atm to bars.

An herbicide is found to contain only \(C, H, N\), and \(C 1\) The complete combustion of a \(100.0-\mathrm{mg}\) sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of \(\mathrm{Cl}\). (a) Determine the percent composition of the substance. (b) Calculate its empirical formula.

Consider the following reaction: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ Imagine that this reaction occurs in a container that has a piston that moves to allow a constant pressure to be maintained when the reaction occurs at constant temperature. (a) What happens to the volume of the container as a result of the reaction? Explain. (b) If the piston is not allowed to move, what happens to the pressure as a result of the reaction? [Sections \(10.3\) and \(10.5]\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.