Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 14, Problem 14.43P

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the formation of BN has to be written.

Concept Introduction:

Balanced equation:

In a chemical equation, the number of atoms or elements that present in both reactant and product sides are equal means this equation is known as balanced equation.

(a)

Expert Solution
Check Mark

Explanation of Solution

Boron nitride is formed with water when, Boron oxide is treated with ammonia.

The balanced equation for this react ion is,

    B2O3(s) + 2NH3(g)2BN(s) + 3H2O(g)

(b)

Interpretation Introduction

Interpretation:

Enthalpy change of a reaction of formation of Boron nitride has to be calculated.

Concept Introduction:

Hess's law:

Hess's law state that the different in enthalpy change of reactant to product is known as enthalpy change of a reaction.

The enthalpy change reactant to product is calculated by subtraction of standard enthalpy of formation (SEF) of reactant from SEF product.

     ΔHreactiono=mΔHproducto-nΔHreactantowhere, ΔHreactiono=enthalpy changeofreaction ΔHProducto=standard enthalpy of formation of reactant ΔHreactanto=standard enthalpy of formation of  product

(b)

Expert Solution
Check Mark

Explanation of Solution

Given,

ΔHf=-254kJ/mol and In3+ are,

Boron nitride is formed with water when, Boron oxide is treated with ammonia.

The balanced equation for this react ion is,

    B2O3(s) + 2NH3(g)2BN(s) + 3H2O(g)

From the standard data, the standard enthalpy of formation of Boron oxide and ammonia are substitute in the below equation we get enthalpy change of a reaction of formation of Boron nitride.

    ={2ΔHf[BN(s)]+3ΔHf[H2O(g)]}-{2ΔHf[B2O3(s)]+2ΔHf[NH3(g)]}=[2(-254kJ/mol)+(3mol)(-241.86kJ/mol)] -[(2mol)(-1272kJ/mol)+(2mol)(-45.0kJ/mol)]=130.322J=1.30×102kJ

Hence, the enthalpy change of a reaction is 1.30×102kJ.

(c)

Interpretation Introduction

Interpretation:

The amount of borax is needed to produce 1.0 kg of BN at  72% yield has to be calculated.

Concept Introduction:

Percentage yield:

Percentage yield is given by,

    %Y=Actual yieldTheoretical yield×100

Mole:

Mole of the solid substance is given by,

    Mole=MassMolarmass

(c)

Expert Solution
Check Mark

Explanation of Solution

The balanced equation for this react ion is,

    B2O3(s) + 2NH3(g)2BN(s) + 3H2O(g)

Produce mass of  BN is 1.0 kg

The amount of borax is needed to produce 1.0 kg of BN at  72% yield is,

    =1.0 kg of BN(1000g1kg)(1molBN24.82gBN)(1molB1molBN)(1molNa2B4O×10H2O4molB)×(381.38g1mol)(100%72.0%)=5.3×103gborax

Hence, the amount of borax is needed to produce 1.0 kg of BN at  72% yield is 5.3×103gborax.

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Chapter 14 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 14 - Prob. 14.11PCh. 14 - Prob. 14.12PCh. 14 - Prob. 14.13PCh. 14 - Prob. 14.14PCh. 14 - Prob. 14.15PCh. 14 - Prob. 14.16PCh. 14 - Prob. 14.17PCh. 14 - Prob. 14.18PCh. 14 - Prob. 14.19PCh. 14 - Prob. 14.20PCh. 14 - Prob. 14.21PCh. 14 - Prob. 14.22PCh. 14 - Prob. 14.23PCh. 14 - Prob. 14.24PCh. 14 - Prob. 14.25PCh. 14 - Prob. 14.26PCh. 14 - Prob. 14.27PCh. 14 - Prob. 14.28PCh. 14 - Prob. 14.29PCh. 14 - Prob. 14.30PCh. 14 - Prob. 14.31PCh. 14 - Prob. 14.32PCh. 14 - Prob. 14.33PCh. 14 - Prob. 14.34PCh. 14 - Prob. 14.35PCh. 14 - Prob. 14.36PCh. 14 - Prob. 14.37PCh. 14 - Prob. 14.38PCh. 14 - Prob. 14.39PCh. 14 - Prob. 14.40PCh. 14 - Prob. 14.41PCh. 14 - Prob. 14.42PCh. 14 - Prob. 14.43PCh. 14 - Prob. 14.44PCh. 14 - Prob. 14.45PCh. 14 - Prob. 14.46PCh. 14 - Give explanations for the large drops in melting...Ch. 14 - Prob. 14.48PCh. 14 - Prob. 14.49PCh. 14 - Prob. 14.50PCh. 14 - Prob. 14.51PCh. 14 - Prob. 14.52PCh. 14 - Prob. 14.53PCh. 14 - Prob. 14.54PCh. 14 - Prob. 14.55PCh. 14 - Prob. 14.56PCh. 14 - Prob. 14.57PCh. 14 - Prob. 14.58PCh. 14 - Prob. 14.59PCh. 14 - Prob. 14.60PCh. 14 - Prob. 14.61PCh. 14 - Prob. 14.62PCh. 14 - Prob. 14.63PCh. 14 - Prob. 14.64PCh. 14 - Prob. 14.65PCh. 14 - Prob. 14.66PCh. 14 - Prob. 14.67PCh. 14 - Prob. 14.68PCh. 14 - Prob. 14.69PCh. 14 - Prob. 14.70PCh. 14 - Prob. 14.71PCh. 14 - Prob. 14.72PCh. 14 - Prob. 14.73PCh. 14 - Prob. 14.74PCh. 14 - Prob. 14.75PCh. 14 - Prob. 14.76PCh. 14 - Prob. 14.77PCh. 14 - Prob. 14.78PCh. 14 - Prob. 14.79PCh. 14 - Prob. 14.80PCh. 14 - Prob. 14.81PCh. 14 - Prob. 14.82PCh. 14 - Prob. 14.83PCh. 14 - Prob. 14.84PCh. 14 - Prob. 14.85PCh. 14 - Prob. 14.86PCh. 14 - Prob. 14.87PCh. 14 - Prob. 14.88PCh. 14 - Prob. 14.89PCh. 14 - Prob. 14.90PCh. 14 - Prob. 14.91PCh. 14 - Prob. 14.92PCh. 14 - Prob. 14.93PCh. 14 - Prob. 14.94PCh. 14 - Prob. 14.95PCh. 14 - Prob. 14.96PCh. 14 - Prob. 14.97PCh. 14 - Prob. 14.98PCh. 14 - Prob. 14.99PCh. 14 - Prob. 14.100PCh. 14 - Prob. 14.101PCh. 14 - Prob. 14.102PCh. 14 - Prob. 14.103PCh. 14 - Prob. 14.104PCh. 14 - Xenon tetrafluoride reacts with antimony...Ch. 14 - Prob. 14.106PCh. 14 - Prob. 14.107PCh. 14 - Prob. 14.108PCh. 14 - Prob. 14.109PCh. 14 - Prob. 14.110PCh. 14 - Prob. 14.111PCh. 14 - Prob. 14.112PCh. 14 - Prob. 14.113PCh. 14 - Prob. 14.114PCh. 14 - Prob. 14.115PCh. 14 - Prob. 14.116PCh. 14 - Prob. 14.117PCh. 14 - Prob. 14.118PCh. 14 - Prob. 14.119PCh. 14 - Prob. 14.120PCh. 14 - Prob. 14.121PCh. 14 - Prob. 14.122PCh. 14 - Prob. 14.123PCh. 14 - Prob. 14.124PCh. 14 - Prob. 14.125PCh. 14 - Prob. 14.126PCh. 14 - Prob. 14.127PCh. 14 - Prob. 14.128PCh. 14 - Prob. 14.129PCh. 14 - Prob. 14.130PCh. 14 - Prob. 14.131PCh. 14 - Prob. 14.132PCh. 14 - Prob. 14.133PCh. 14 - Prob. 14.134PCh. 14 - Hydrogen peroxide can act as either an oxidizing...
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