Interpretation Introduction To determine: The volume of the gas sample. Explanation Given: M1 ass of oxygen, m o 2 = 1.15 g Mass of nitrogen, m N 2 = 1.55 g Pressure at STP, P = 1 atm Temperature at STP, T = 273 K Explanation: The volumes of the individual gases, oxygen and nitrogen, are determined by using the ideal gas law : V = n R T P In order to calculate the volume, the masses of oxygen and nitrogen are converted into their moles. Convert the masses of oxygen and nitrogen into their moles: Number of moles ( n ) = Mass of ( m ) Molar mass ( M ) Substitute the mass of oxygen, m o 2 = 1.15 g , and molar mass of oxygen, M = 32 g/mol , in the above equation to get Number of moles of oxygen gas ( n O 2 ) = 1 .15 g 32 g / mol = 0 .035 mol Similarly, substitute the mass of nitrogen, m N 2 = 1.55 g , and molar mass of nitrogen, M = 28 g/mol , in the above equation to get Number of moles of nitrogen gas ( n N 2 ) = 1 .55 g 28 g / mol = 0 .055 mol The volume of oxygen is given by V O 2 = n O 2 R T P Substitute n O 2 = 0.035 mol , P = 1 atm , R = 0

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

1st Edition

ISBN: 9780321955517

Author: Nivaldo J. Tro

Publisher: PEARSON

See similar textbooks

Videos

Question

Chapter 5, Problem 9SAQ

Interpretation Introduction

To determine: The volume of the gas sample.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5, Problem 8SAQ

Chapter 5, Problem 10SAQ

Students have asked these similar questions

Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NO(g) +Cl₂ (g) = 2NOC1 (g) AGº = -41. kJ Now suppose a reaction vessel is filled with 8.90 atm of chlorine (C12) and 5.71 atm of nitrosyl chloride (NOC1) at 1075. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of NOCI tend to rise or fall? x10 fall Is it possible to reverse this tendency by adding NO? In other words, if you said the pressure of NOCI will tend to rise, can that be changed to a tendency to fall by adding NO? Similarly, if you said the pressure of NOCI will tend to fall, can that be changed to a tendency to rise by adding NO? yes no If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO needed to reverse it. Round your answer to 2 significant digits. atm ☑ 18 Ar

Identifying the major species in weak acid or weak base equilibria The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HCN is a weak acid. acids: 0.29 mol of NaOH is added to 1.0 L of a 1.2M HCN solution. bases: ☑ other: 0.09 mol of HCl is added to acids: 1.0 L of a solution that is bases: 0.3M in both HCN and KCN. other: 0,0,... ? 00. 18 Ar 日

Identifying the major species in weak acid or weak base equilibria The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HF is a weak acid. acids: 0.2 mol of KOH is added to 1.0 L of a 0.5 M HF solution. bases: Х other: ☐ acids: 0.10 mol of HI is added to 1.0 L of a solution that is 1.4M in both HF and NaF. bases: other: ☐ 0,0,... ด ? 18 Ar

Chapter 5 Solutions

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

Ch. 5 - Prob. 1SAQ Ch. 5 - Prob. 2SAQ Ch. 5 - Prob. 3SAQ Ch. 5 - Prob. 4SAQ Ch. 5 - Prob. 5SAQ Ch. 5 - Prob. 6SAQ Ch. 5 - Prob. 7SAQ Ch. 5 - Prob. 8SAQ Ch. 5 - Q9. A gas sample at STP contains 1.15 g oxygen and...Ch. 5 - Prob. 10SAQ

Ch. 5 - Prob. 11SAQ Ch. 5 - Prob. 12SAQ Ch. 5 - Prob. 13SAQ Ch. 5 - Prob. 14SAQ Ch. 5 - Prob. 15SAQ Ch. 5 - Prob. 1E Ch. 5 - Prob. 2E Ch. 5 - Prob. 3E Ch. 5 - Prob. 4E Ch. 5 - Prob. 5E Ch. 5 - Prob. 6E Ch. 5 - Prob. 7E Ch. 5 - Prob. 8E Ch. 5 - Prob. 9E Ch. 5 - Prob. 10E Ch. 5 - Prob. 11E Ch. 5 - Prob. 12E Ch. 5 - Prob. 13E Ch. 5 - Prob. 14E Ch. 5 - Prob. 15E Ch. 5 - Prob. 16E Ch. 5 - Prob. 17E Ch. 5 - Prob. 18E Ch. 5 - Prob. 19E Ch. 5 - Prob. 20E Ch. 5 - Prob. 21E Ch. 5 - Prob. 22E Ch. 5 - Prob. 23E Ch. 5 - Prob. 24E Ch. 5 - Prob. 25E Ch. 5 - Prob. 26E Ch. 5 - Prob. 27E Ch. 5 - Prob. 28E Ch. 5 - 29. Given a barometric pressure of 762.4 mmHg,...Ch. 5 - Prob. 30E Ch. 5 - Prob. 31E Ch. 5 - Prob. 32E Ch. 5 - Prob. 33E Ch. 5 - Prob. 34E Ch. 5 - Prob. 35E Ch. 5 - Prob. 36E Ch. 5 - Prob. 37E Ch. 5 - Prob. 38E Ch. 5 - Prob. 39E Ch. 5 - Prob. 40E Ch. 5 - Prob. 41E Ch. 5 - Prob. 42E Ch. 5 - Prob. 43E Ch. 5 - Prob. 44E Ch. 5 - Prob. 45E Ch. 5 - Prob. 46E Ch. 5 - Prob. 47E Ch. 5 - Prob. 48E Ch. 5 - Prob. 49E Ch. 5 - Prob. 50E Ch. 5 - Prob. 51E Ch. 5 - Prob. 52E Ch. 5 - Prob. 53E Ch. 5 - Prob. 54E Ch. 5 - Prob. 55E Ch. 5 - Prob. 56E Ch. 5 - Prob. 57E Ch. 5 - Prob. 58E Ch. 5 - Prob. 59E Ch. 5 - Prob. 60E Ch. 5 - Prob. 61E Ch. 5 - Prob. 62E Ch. 5 - Prob. 63E Ch. 5 - 64. A 275-mL flask contains pure helium at a...Ch. 5 - Prob. 65E Ch. 5 - Prob. 66E Ch. 5 - Prob. 67E Ch. 5 - Prob. 68E Ch. 5 - Prob. 69E Ch. 5 - Prob. 70E Ch. 5 - Prob. 71E Ch. 5 - Prob. 72E Ch. 5 - Prob. 73E Ch. 5 - Prob. 74E Ch. 5 - Prob. 75E Ch. 5 - Prob. 76E Ch. 5 - Prob. 77E Ch. 5 - 78. Ozone is depleted in the stratosphere by...Ch. 5 - Prob. 79E Ch. 5 - Prob. 80E Ch. 5 - Prob. 81E Ch. 5 - Prob. 82E Ch. 5 - Prob. 83E Ch. 5 - Prob. 84E Ch. 5 - Prob. 85E Ch. 5 - Prob. 86E Ch. 5 - Prob. 87E Ch. 5 - Prob. 88E Ch. 5 - Prob. 89E Ch. 5 - Prob. 90E Ch. 5 - Prob. 91E Ch. 5 - Prob. 92E Ch. 5 - Prob. 93E Ch. 5 - Prob. 94E Ch. 5 - 95. Modern pennies are composed of zinc coated...Ch. 5 - Prob. 96E Ch. 5 - Prob. 97E Ch. 5 - Prob. 98E Ch. 5 - Prob. 99E Ch. 5 - Prob. 100E Ch. 5 - Prob. 101E Ch. 5 - Prob. 102E Ch. 5 - Prob. 103E Ch. 5 - Prob. 104E Ch. 5 - Prob. 105E Ch. 5 - Prob. 106E Ch. 5 - Prob. 107E Ch. 5 - Prob. 108E Ch. 5 - Prob. 109E Ch. 5 - Prob. 110E Ch. 5 - Prob. 111E Ch. 5 - Prob. 112E Ch. 5 - Prob. 113E Ch. 5 - Prob. 114E Ch. 5 - Prob. 115E Ch. 5 - Prob. 116E Ch. 5 - Prob. 117E Ch. 5 - Prob. 118E Ch. 5 - Prob. 119E Ch. 5 - Prob. 120E Ch. 5 - Prob. 121E Ch. 5 - Prob. 122E Ch. 5 - Prob. 123E Ch. 5 - Prob. 124E Ch. 5 - Prob. 125E Ch. 5 - Prob. 126E Ch. 5 - Prob. 127E Ch. 5 - Prob. 128E Ch. 5 - Prob. 129E Ch. 5 - Prob. 130E Ch. 5 - Prob. 131E Ch. 5 - Prob. 132E Ch. 5 - Prob. 133E Ch. 5 - Prob. 134E Ch. 5 - Prob. 135E Ch. 5 - Prob. 136E Ch. 5 - Prob. 137E Ch. 5 - Prob. 138E Ch. 5 - Prob. 139E Ch. 5 - Prob. 140E Ch. 5 - Prob. 141E Ch. 5 - Prob. 142E Ch. 5 - Prob. 143E Ch. 5 - Prob. 144E Ch. 5 - Prob. 145E Ch. 5 - Prob. 146E Ch. 5 - Prob. 147E

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.