Explanation Given: Mass of lithium is 2 .50 g . Volume of the solution is 1 .500 L . Formula used: Number of moles calculated by the formula, n = m M Where, n is the number of moles. m is the mass of the molecule. M is the molar mass of the molecule. M = n V Where, M is the molarity. n is the number of moles. V is the volume of the solution. Calculation: The molar mass of Li 2 O is, = ( 2×6 .94 g/mol ) + ( 16 g/mol ) = 29 .88 g/mol Number of moles calculated by the formula, n = m M Where, n is the number of moles. m is the mass of the molecule. M is the molar mass of the molecule. Substitute the value of mass of Li 2 O and molar mass of Li 2 O in the above formula. n = 2 .50 g 29 .88 g/mol = 0 .083 mol Molarity is calculated by the following formula, M = n V Where, M is the molarity. n is the number of moles. V is the volume of the solution. Substitute the value of number of moles of Li 2 O and volume of the solution in the above expression

Laboratory Experiments For Chemistry: The Central Science, Si Edition

14th Edition

ISBN: 9781292221335

Author: Theodore E. Brown

Publisher: PEARSON

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Chapter 16, Problem 104AE

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To determine: The value of pH for the given reaction.

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Chapter 16, Problem 103AE

Chapter 16, Problem 105AE

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Chapter 16 Solutions

Laboratory Experiments For Chemistry: The Central Science, Si Edition

Ch. 16.2 - Practice Exercise 1 Consider the following...Ch. 16.2 - Prob. 16.1.2PE Ch. 16.2 - Prob. 16.2.1PE Ch. 16.2 - Practice Exercise 2 When lithium oxide (Li2O) is...Ch. 16.2 - Based on information in Figure 16.4, place the...Ch. 16.2 - Practice Exercise 2 For each reaction, use Figure...Ch. 16.3 - Practice Exercise 1 In a certain acidic solution...Ch. 16.3 - Practice Exercise 2 Indicate whether solutions...Ch. 16.3 - Prob. 16.5.1PE Ch. 16.3 - Prob. 16.5.2PE

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