Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 20, Problem 117IP

(a)

Interpretation Introduction

Interpretation: The reaction between In(CH3)3 and PH3 at 900K is given. The answers are to be stated for the given options.

Concept introduction: The number of moles is calculated using ideal gas law,

PV=nRTn=PVRT

To determine: The mass of InP when 2.56L In(CH3)3 at 2.00atm is allowed to react with 1.38L PH3 at 3.00atm (assuming the reaction has 87% yield).

(a)

Expert Solution
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Explanation of Solution

Explanation

Given

Temperature is 900K .

Pressure of In(CH3)3 is 2.00atm .

Volume of In(CH3)3 is 2.56L .

Pressure of PH3 is 3.00atm .

Volume of PH3 is 1.38L .

Formula

Number of moles is calculated as,

PV=nRTn=PVRT (1)

Where,

  • P is the total pressure.
  • V is the volume.
  • n is the total moles.
  • R is the universal gas constant (0.08206Latm/Kmol) .
  • T is the absolute temperature.

Substitute the values of P,V,R and T for PH3 in the above equation.

nPH3=PVRT=2.00atm×1.38L(0.08206Latm/Kmol)(900K)=0.056mol_

Substitute the values of P,V,R and T for In(CH3)3 in equation (1).

nIn(CH3)3=PVRT=2.00atm×2.56L(0.08206Latm/Kmol)(900K)=0.0693mol_

To determine the mass of InP is 7.1g_ .

Moles of In(CH3)3 is 0.0693mol .

Moles of PH3 is 0.056mol .

The moles of PH3 is less than the moles of In(CH3)3 . Hence, PH3 will acts as limiting reagent.

Formula

Mass of InP is calculated using the formula,

MassofInP=MolesofPH3×1molInP1molPH3×MolarmassofInP1molInP

Substitute the values of moles of PH3 and the molar mass of InP in the above equation.

MassofInP=MolesofPH3×1molInP1molPH3×MolarmassofInP1molInP=0.056molPH3×1molInP1molPH3×145.791molInP=8.164g

Since, it is assumed that the reaction has 87% yield. Therefore mass of InP formed is,

8.164g×87100=7.1g_

(b)

Interpretation Introduction

Interpretation: The reaction between In(CH3)3 and PH3 at 900K is given. The answers are to be stated for the given options.

Concept introduction: The number of moles is calculated using ideal gas law,

PV=nRTn=PVRT

To determine: The wavelength of light if 2.03×1019J of energy is emitted by light; if this wavelength is visible to human eye.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given

Energy is 2.03×1019J .

Formula

The wavelength of light is calculated using the formula,

E=hcλλ=hcE

Where,

  • λ is the wavelength.
  • c is the velocity of light (3×108m/s)
  • E is the energy.
  • h is the Plank’s constant (6.626×1034J/s) .

Substitute the values of E,h and c in the above equation.

λ=hcE=(6.626×1034J/s)(3×108m/s)2.03×1019J=979×109m

The conversion of meter (m) into nanometer (nm) is done as,

1m=109nm

Hence,

The conversion of 979×109m into nanometer is,

979×109m=(979×109×109)nm=979nm_

This wavelength is not visible to human eye.

The calculated value of wavelength of light is 979nm .

This wavelength of light belongs to infrared region. Hence, this wavelength is not visible to human eye.

(c)

Interpretation Introduction

Interpretation: The reaction between In(CH3)3 and PH3 at 900K is given. The answers are to be stated for the given options.

Concept introduction: The number of moles is calculated using ideal gas law,

PV=nRTn=PVRT

To determine: If InP becomes n-type or p-type when small number of phosphorous atoms are replaced by the atoms of electronic configuration [Kr]5s24d105p4 .

(c)

Expert Solution
Check Mark

Explanation of Solution

The material InP becomes n-type when small number of phosphorous atoms are replaced by the atoms of electronic configuration [Kr]5s24d105p4 .

The given electronic configuration [Kr]5s24d105p4 belongs to group 15th . This is the electronic configuration of arsenic. Arsenic is the pentavalent element. Hence, InP becomes n-type.

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