Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 3, Problem 3.9E

(a)

Interpretation Introduction

Interpretation:

Molecular formula for unknown compound has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  PV=nRT        (1)

Here,

P is pressure of the gas.

V is volume of gas.

n denotes moles of gas.

R is gas constant.

T is temperature of gas.

(a)

Expert Solution
Check Mark

Answer to Problem 3.9E

Mass of NaN3 required to fill air bag with nitrogen is 138 g.

Explanation of Solution

Conversion factor to convert 25 °C to Kelvin is as follows:

  T(K)=temperatureincelsius+273.15=25 °C+273.15=298.15K

Rearrange equation (1) to determine value of n for unknown gas as follows:

  n=PVRT        (2)

Substitute 1.81 atm for P, 200 mL for V, 8.20574×102 LatmK1mol1 for R, and 298.15K for T in equation (2).

  n=((1.81 atm)(200 mL)(8.20574×102 LatmK1mol1)(298.15K))(103 L 1 mL)=1.48×102 mol

Number of moles of nitrogen present in 0.414 g can be calculated as follows:

  Moles=0.414 g14 gmol1=0.02957 mol

Number of moles of hydrogen gas present in 0.0591 g can be calculated as follows:

  Moles=0.0591 g1 gmol1=0.0591 mol

Ratio of nitrogen to hydrogen can be calculated as follows:

  Ratio(N:H)=0.02957 mol0.0591 mol=12

Therefore formula unit is NH2. Molar mass of formula unit can be calculated as follows:

  Molar mass=1(14 gmol1)+2(1 gmol1)=16 gmol1

Total molar mass of gas can be calculated as follows:

  Molar mass=MassMoles=0.473 g1.48×102 mol32 gmol1

Therefore, number of formula unit of NH2 present in unknown gas is 2. Hence molecular formula for unknown gas is N2H4.

(b)

Interpretation Introduction

Interpretation:

Lewis structure of N2H4 has to be determined.

Concept Introduction:

Lewis structure represents covalent bonds and describes valence electrons configuration of atoms. The covalent bonds are depicted by lines and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

  • The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
  • Total valence electrons are estimated for each atom.
  • single bond is first placed between each atom pair.
  • The electrons left can be allocated as unshared electron pairs or as multiple bonds around symbol of element to satisfy the octet (or duplet) for each atom.
  • Add charge on overall structure in case of polyatomic cation or anion.

(b)

Expert Solution
Check Mark

Explanation of Solution

The molecule N2H4 consists of two N and four H atoms.

Electronic configuration of N is [He]2s22p3. Thus, it possesses 5 valence electrons.

Electronic configuration of H is 1s1. Thus, it possesses 1 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in N2H4. It is calculated as follows:

  Total valence electrons=2(5)+4(1)=14(7 pairs)

Skeleton structure N2H4 has four single bonds between hydrogen and nitrogen that comprise 8 electrons. Also, one single bond between nitrogen and nitrogen that comprise 2 electrons. The electrons left to be allocated are determined as follows:

  Remaining electrons=1410=4(2 pairs)

Hence, 4 electrons are allocated as 2 lone pairs on each nitrogen atom. The Lewis structure of N2H4 is as follows:

Chemical Principles: The Quest for Insight, Chapter 3, Problem 3.9E

(c)

Interpretation Introduction

Interpretation:

Amount of nitrogen gas effused through small hole in 25.0 min has to be determined.

Concept Introduction:

Effusion is explained as movement of the gas molecule through small hole. Diffusion can be explained as one gas molecule mix with another gas molecule by random motion.

(c)

Expert Solution
Check Mark

Answer to Problem 3.9E

Amount of nitrogen gas effused through small hole in 25.0 min is 0.42 mmol.

Explanation of Solution

The expression for molar masses of N2H4, NH3 and their effusion rate is as follows:

  Amount of N2H4 passedTime taken by N2H4Amount of NH3 passedTime taken by NH3=MNH3MN2H4        (3)

Rearrange equation (3) to calculate amount of NH3 as follows:

  Amount of N2H4 passed=((Time taken by N2H4)(MNH3MN2H4)(Amount of NH3 passedTime taken by NH3))        (4)

Substitute 25.0 min time taken by N2H4, 15.0 min time taken by NH3, 0.35 mmol amount of NH3, 17.03 g/mol for molar mass of NH3, and 32.05 g/mol for molar mass of N2H4 in equation (4).

  Amount of N2H4 passed=((25.0 min)(17.03 g/mol32.05 g/mol)(0.35 mmol15.0 min))=0.42 mmol

Hence amount of nitrogen gas effused through small hole in 25.0 min is 0.42 mmol.

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Chapter 3 Solutions

Chemical Principles: The Quest for Insight

Ch. 3 - Prob. 3A.5ECh. 3 - Prob. 3A.6ECh. 3 - Prob. 3A.7ECh. 3 - Prob. 3A.8ECh. 3 - Prob. 3A.9ECh. 3 - Prob. 3A.10ECh. 3 - Prob. 3B.1ASTCh. 3 - Prob. 3B.1BSTCh. 3 - Prob. 3B.2ASTCh. 3 - Prob. 3B.2BSTCh. 3 - Prob. 3B.3ASTCh. 3 - Prob. 3B.3BSTCh. 3 - Prob. 3B.4ASTCh. 3 - Prob. 3B.4BSTCh. 3 - Prob. 3B.5ASTCh. 3 - Prob. 3B.5BSTCh. 3 - Prob. 3B.6ASTCh. 3 - Prob. 3B.6BSTCh. 3 - Prob. 3B.7ASTCh. 3 - Prob. 3B.7BSTCh. 3 - Prob. 3B.8ASTCh. 3 - Prob. 3B.8BSTCh. 3 - Prob. 3B.1ECh. 3 - Prob. 3B.2ECh. 3 - Prob. 3B.5ECh. 3 - Prob. 3B.6ECh. 3 - Prob. 3B.9ECh. 3 - Prob. 3B.10ECh. 3 - Prob. 3B.11ECh. 3 - Prob. 3B.12ECh. 3 - Prob. 3B.13ECh. 3 - Prob. 3B.14ECh. 3 - Prob. 3B.15ECh. 3 - Prob. 3B.16ECh. 3 - Prob. 3B.17ECh. 3 - Prob. 3B.18ECh. 3 - Prob. 3B.19ECh. 3 - Prob. 3B.20ECh. 3 - Prob. 3B.21ECh. 3 - Prob. 3B.22ECh. 3 - Prob. 3B.23ECh. 3 - Prob. 3B.24ECh. 3 - Prob. 3B.25ECh. 3 - Prob. 3B.26ECh. 3 - Prob. 3B.27ECh. 3 - Prob. 3B.28ECh. 3 - Prob. 3B.29ECh. 3 - Prob. 3B.30ECh. 3 - Prob. 3B.31ECh. 3 - Prob. 3B.32ECh. 3 - 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Prob. 3G.7ECh. 3 - Prob. 3G.8ECh. 3 - Prob. 3G.9ECh. 3 - Prob. 3G.10ECh. 3 - Prob. 3G.11ECh. 3 - Prob. 3G.12ECh. 3 - Prob. 3G.13ECh. 3 - Prob. 3G.14ECh. 3 - Prob. 3G.15ECh. 3 - Prob. 3G.16ECh. 3 - Prob. 3G.17ECh. 3 - Prob. 3G.18ECh. 3 - Prob. 3H.1ASTCh. 3 - Prob. 3H.1BSTCh. 3 - Prob. 3H.2ASTCh. 3 - Prob. 3H.2BSTCh. 3 - Prob. 3H.3ASTCh. 3 - Prob. 3H.3BSTCh. 3 - Prob. 3H.4ASTCh. 3 - Prob. 3H.4BSTCh. 3 - Prob. 3H.5ASTCh. 3 - Prob. 3H.5BSTCh. 3 - Prob. 3H.1ECh. 3 - Prob. 3H.2ECh. 3 - Prob. 3H.3ECh. 3 - Prob. 3H.4ECh. 3 - Prob. 3H.5ECh. 3 - Prob. 3H.6ECh. 3 - Prob. 3H.7ECh. 3 - Prob. 3H.8ECh. 3 - Prob. 3H.9ECh. 3 - Prob. 3H.10ECh. 3 - Prob. 3H.11ECh. 3 - Prob. 3H.12ECh. 3 - Prob. 3H.13ECh. 3 - Prob. 3H.14ECh. 3 - Prob. 3H.15ECh. 3 - Prob. 3H.16ECh. 3 - Prob. 3H.17ECh. 3 - Prob. 3H.19ECh. 3 - Prob. 3H.20ECh. 3 - Prob. 3H.23ECh. 3 - Prob. 3H.24ECh. 3 - Prob. 3H.25ECh. 3 - Prob. 3H.26ECh. 3 - Prob. 3H.27ECh. 3 - Prob. 3H.28ECh. 3 - Prob. 3H.29ECh. 3 - Prob. 3H.30ECh. 3 - Prob. 3H.31ECh. 3 - Prob. 3H.32ECh. 3 - Prob. 3H.33ECh. 3 - Prob. 3H.34ECh. 3 - Prob. 3H.35ECh. 3 - Prob. 3H.36ECh. 3 - Prob. 3I.1ASTCh. 3 - Prob. 3I.1BSTCh. 3 - Prob. 3I.2ASTCh. 3 - Prob. 3I.2BSTCh. 3 - Prob. 3I.3ASTCh. 3 - Prob. 3I.3BSTCh. 3 - Prob. 3I.4ASTCh. 3 - Prob. 3I.4BSTCh. 3 - Prob. 3I.1ECh. 3 - Prob. 3I.2ECh. 3 - Prob. 3I.3ECh. 3 - Prob. 3I.4ECh. 3 - Prob. 3I.5ECh. 3 - Prob. 3I.6ECh. 3 - Prob. 3I.7ECh. 3 - Prob. 3I.8ECh. 3 - Prob. 3I.11ECh. 3 - Prob. 3I.12ECh. 3 - Prob. 3I.13ECh. 3 - Prob. 3I.14ECh. 3 - Prob. 3I.15ECh. 3 - Prob. 3I.16ECh. 3 - Prob. 3J.1ASTCh. 3 - Prob. 3J.1BSTCh. 3 - Prob. 3J.2ASTCh. 3 - Prob. 3J.2BSTCh. 3 - Prob. 3J.3ASTCh. 3 - Prob. 3J.3BSTCh. 3 - Prob. 3J.1ECh. 3 - Prob. 3J.2ECh. 3 - Prob. 3J.3ECh. 3 - Prob. 3J.4ECh. 3 - Prob. 3J.5ECh. 3 - Prob. 3J.6ECh. 3 - Prob. 3J.7ECh. 3 - Prob. 3J.8ECh. 3 - Prob. 3J.9ECh. 3 - Prob. 3J.10ECh. 3 - Prob. 3J.11ECh. 3 - Prob. 3J.12ECh. 3 - Prob. 3J.13ECh. 3 - Prob. 3J.14ECh. 3 - Prob. 3J.15ECh. 3 - Prob. 3J.16ECh. 3 - Prob. 3.1ECh. 3 - Prob. 3.2ECh. 3 - Prob. 3.3ECh. 3 - Prob. 3.4ECh. 3 - Prob. 3.5ECh. 3 - Prob. 3.6ECh. 3 - Prob. 3.7ECh. 3 - Prob. 3.8ECh. 3 - Prob. 3.9ECh. 3 - Prob. 3.10ECh. 3 - Prob. 3.11ECh. 3 - Prob. 3.12ECh. 3 - Prob. 3.13ECh. 3 - Prob. 3.15ECh. 3 - Prob. 3.18ECh. 3 - Prob. 3.19ECh. 3 - Prob. 3.23ECh. 3 - Prob. 3.24ECh. 3 - Prob. 3.25ECh. 3 - Prob. 3.26ECh. 3 - Prob. 3.27ECh. 3 - Prob. 3.29ECh. 3 - Prob. 3.31ECh. 3 - Prob. 3.32ECh. 3 - Prob. 3.35ECh. 3 - Prob. 3.36ECh. 3 - Prob. 3.37ECh. 3 - Prob. 3.38ECh. 3 - Prob. 3.40ECh. 3 - Prob. 3.41ECh. 3 - Prob. 3.42ECh. 3 - Prob. 3.45ECh. 3 - Prob. 3.47ECh. 3 - Prob. 3.49ECh. 3 - Prob. 3.50ECh. 3 - Prob. 3.51ECh. 3 - Prob. 3.53ECh. 3 - Prob. 3.54ECh. 3 - Prob. 3.55ECh. 3 - Prob. 3.56ECh. 3 - Prob. 3.57ECh. 3 - Prob. 3.58ECh. 3 - Prob. 3.59ECh. 3 - Prob. 3.60ECh. 3 - Prob. 3.61ECh. 3 - Prob. 3.62ECh. 3 - Prob. 3.63ECh. 3 - Prob. 3.64ECh. 3 - Prob. 3.65ECh. 3 - Prob. 3.66ECh. 3 - Prob. 3.67ECh. 3 - Prob. 3.68E
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