Principles of General, Organic, Biological Chemistry
Principles of General, Organic, Biological Chemistry
2nd Edition
ISBN: 9780073511191
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
Question
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Chapter 7, Problem 7.73AP

(a)

Interpretation Introduction

Interpretation:

The grams of sodium nitrate present have to be calculated.

Concept Introduction:

Molarity:  Molarity is defined as the mass of solute in one liter of solution.  Molarity is the preferred concentration unit for stoichiometry calculations.  The formula is,

  Molarity=Massofsolute(inmoles)Volumeofsolution(inlitres)

(a)

Expert Solution
Check Mark

Answer to Problem 7.73AP

The grams of sodium nitrate is 3.23g.

Explanation of Solution

Given,

Volume of solution=150mL

Molarity of solution=0.25M

Milliliters is converted into liters,

  L=150mL×1L1000mLL=0.150L

The moles of sodium nitrate is calculated as,

  Molarity=Massofsolute(inmoles)Volumeofsolution(inlitres)0.25M=x0.150Lx=(0.150L)(0.25molL)x=0.0375mole

The moles of sodium nitrate is 0.038mol.

Moles to grams is converted as,

  Grams=0.038mol NaNO3×85.00g NaNO31moleNaNO3Grams=3.23g

The grams of sodium nitrate is 3.23g.

(b)

Interpretation Introduction

Interpretation:

The grams of nitric acid present have to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 7.73AP

The grams of nitric acid is 5.7g.

Explanation of Solution

Given,

Volume of solution=45mL

Molarity of solution=2.0M

Milliliters is converted into liters,

  L=45mL×1L1000mLL=0.045L

The moles of nitric acid is calculated as,

  Molarity=Massofsolute(inmoles)Volumeofsolution(inlitres)2.0M=x0.045Lx=(0.045L)(2.0molL)x=0.09mole

The moles of nitric acid is 0.09mol.

Moles to grams is converted as,

  Grams=0.09mol HNO3×63.02g HNO31moleHNO3Grams=5.6718g

The grams of nitric acid is 5.7g.

(c)

Interpretation Introduction

Interpretation:

The grams of hydrochloric acid present have to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 7.73AP

The grams of hydrochloric acid is 138.548g.

Explanation of Solution

Given,

Volume of solution=2.5L

Molarity of solution=1.5M

The moles of hydrochloric acid is calculated as,

  Molarity=Massofsolute(inmoles)Volumeofsolution(inlitres)1.5M=x2.5Lx=(2.5L)(1.5molL)x=3.75mole

The moles of hydrochloric acid is 3.8mol.

Moles to grams is converted as,

  Grams=3.8mol HCl×36.46g HCl1moleHClGrams=138.548g

The grams of hydrochloric acid is 138.548g.

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Chapter 7 Solutions

Principles of General, Organic, Biological Chemistry

Ch. 7.4 - Why does a soft drink become flat faster when it...Ch. 7.4 - Predict the effect each change has on the...Ch. 7.5 - Prob. 7.13PCh. 7.5 - Prob. 7.14PCh. 7.5 - Prob. 7.15PCh. 7.5 - A drink sold in a health food store contains 0.50%...Ch. 7.5 - Prob. 7.17PCh. 7.5 - Prob. 7.18PCh. 7.5 - Prob. 7.19PCh. 7.6 - Prob. 7.20PCh. 7.6 - Prob. 7.21PCh. 7.6 - Prob. 7.22PCh. 7.6 - Prob. 7.23PCh. 7.6 - Prob. 7.24PCh. 7.7 - Prob. 7.25PCh. 7.7 - Prob. 7.26PCh. 7.7 - Prob. 7.27PCh. 7.8 - Which solution in each pair exerts the greater...Ch. 7.8 - Describe the process that occurs when a 1.0 M NaCl...Ch. 7.8 - Prob. 7.30PCh. 7 - Prob. 7.31UKCCh. 7 - Prob. 7.32UKCCh. 7 - Prob. 7.33UKCCh. 7 - Prob. 7.34UKCCh. 7 - Prob. 7.35UKCCh. 7 - Prob. 7.36UKCCh. 7 - Prob. 7.37UKCCh. 7 - Prob. 7.38UKCCh. 7 - Prob. 7.41UKCCh. 7 - Prob. 7.42UKCCh. 7 - Prob. 7.43APCh. 7 - Prob. 7.44APCh. 7 - Prob. 7.45APCh. 7 - Prob. 7.46APCh. 7 - Prob. 7.47APCh. 7 - Prob. 7.48APCh. 7 - Prob. 7.49APCh. 7 - Prob. 7.50APCh. 7 - Prob. 7.51APCh. 7 - Prob. 7.52APCh. 7 - Prob. 7.53APCh. 7 - Prob. 7.54APCh. 7 - Prob. 7.55APCh. 7 - Prob. 7.56APCh. 7 - Prob. 7.57APCh. 7 - Prob. 7.58APCh. 7 - Prob. 7.59APCh. 7 - Prob. 7.60APCh. 7 - Prob. 7.61APCh. 7 - Prob. 7.62APCh. 7 - Prob. 7.63APCh. 7 - Prob. 7.64APCh. 7 - Prob. 7.65APCh. 7 - Prob. 7.66APCh. 7 - Prob. 7.67APCh. 7 - Prob. 7.68APCh. 7 - Prob. 7.69APCh. 7 - Prob. 7.70APCh. 7 - Prob. 7.71APCh. 7 - Prob. 7.72APCh. 7 - Prob. 7.73APCh. 7 - Prob. 7.74APCh. 7 - Prob. 7.75APCh. 7 - Prob. 7.76APCh. 7 - Prob. 7.77APCh. 7 - Prob. 7.78APCh. 7 - Prob. 7.79APCh. 7 - Prob. 7.80APCh. 7 - Prob. 7.81APCh. 7 - Prob. 7.82APCh. 7 - Prob. 7.83APCh. 7 - Prob. 7.84APCh. 7 - Prob. 7.85APCh. 7 - Prob. 7.86APCh. 7 - Prob. 7.87APCh. 7 - Prob. 7.88APCh. 7 - Prob. 7.89APCh. 7 - Prob. 7.90APCh. 7 - If the concentration of glucose in the blood is 90...Ch. 7 - Prob. 7.92APCh. 7 - Prob. 7.93APCh. 7 - Prob. 7.94APCh. 7 - Prob. 7.95APCh. 7 - Prob. 7.96APCh. 7 - Prob. 7.97APCh. 7 - Prob. 7.98APCh. 7 - Prob. 7.99APCh. 7 - Prob. 7.100APCh. 7 - Prob. 7.101APCh. 7 - Prob. 7.102APCh. 7 - Prob. 7.103CPCh. 7 - Prob. 7.104CP
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