Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
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
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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

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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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