Living by Chemistry
Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter U4, Problem C13.4RE

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given chemical reaction of nitrogen gas and hydrogen gas needs to be determined.

Concept Introduction:

The chemical reaction must be balanced such that it agrees with the law of mass conservation. A balanced chemical equation arises when the number of atoms or ions of the reactant side equals the number of atoms or ions of the product side.

(a)

Expert Solution
Check Mark

Answer to Problem C13.4RE

The balanced chemical equation is:

  N2(g)  +  2H2(g)    N2H4(g)

Explanation of Solution

The unbalanced chemical reaction is as follows:

  N2(g)  +  H2(g)    N2H4(g)

To balance the above reaction, give coefficient 2 to H2 thus, the balanced chemical reaction will be as follows:

  N2(g)  +  2H2(g)    N2H4(g)

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given chemical reaction of potassium nitrate and potassium metal needs to be determined.

Concept Introduction:

The chemical reaction must be balanced such that it agrees with the law of mass conservation. A balanced chemical equation arises when the number of atoms or ions of the reactant side equals the number of atoms or ions of the product side.

(b)

Expert Solution
Check Mark

Answer to Problem C13.4RE

The balanced chemical equation is:

  2KNO3(s)  +  10K(s)    6K2O(s)  +  N2(g)

Explanation of Solution

The unbalanced chemical reaction is as follows:

  KNO3(s)  +  K(s)    K2O(s)  +  N2(g)

To balance the above reaction, give coefficient 2 to KNO3, 10 to K and 6 to K2O, balanced chemical reaction will be as follows:

  2KNO3(s)  +  10K(s)    6K2O(s)  +  N2(g)

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given chemical reaction of aqueous sulfuric acid and aqueous sodium cyanide needs to be determined.

Concept Introduction:

The chemical reaction must be balanced such that it agrees with the law of mass conservation. A balanced chemical equation arises when the number of atoms or ions of the reactant side equals the number of atoms or ions of the product side.

(c)

Expert Solution
Check Mark

Answer to Problem C13.4RE

The balanced chemical equation is:

  H2SO4(aq)  +  2NaCN(aq)    2HCN(g)  +  Na2SO4(aq)

Explanation of Solution

The unbalanced chemical reaction is as follows:

  H2SO4(aq)  +  NaCN(aq)    HCN(g)  +  Na2SO4(aq)

To balance the above reaction, give coefficient 2 to NaCN and 2 to HCN, balanced chemical reaction will be as follows:

  H2SO4(aq)  +  2NaCN(aq)    2HCN(g)  +  Na2SO4(aq)

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given chemical reaction of phosphoric acid and calcium hydroxide needs to be determined.

Concept Introduction:

The chemical reaction must be balanced such that it agrees with the law of mass conservation. A balanced chemical equation arises when the number of atoms or ions of the reactant side equals the number of atoms or ions of the product side.

(d)

Expert Solution
Check Mark

Answer to Problem C13.4RE

The balanced chemical equation is:

  2H3PO4(aq)  +  3Ca(OH)2(aq)    Ca3(PO4)2(aq)  +  6H2O(l)

Explanation of Solution

The unbalanced chemical reaction is as follows:

  H3PO4(aq)  +  Ca(OH)2(aq)    Ca3(PO4)2(aq)  +  H2O(l)

To balance the above reaction, give coefficient 2 to H3PO4, 3 to Ca(OH)2 and 6 to H2O, thus, the balanced chemical reaction will be as follows:

  2H3PO4(aq)  +  3Ca(OH)2(aq)    Ca3(PO4)2(aq)  +  6H2O(l)

(e)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given chemical reaction of propane and oxygen gas needs to be determined.

Concept Introduction:

The chemical reaction must be balanced such that it agrees with the law of mass conservation. A balanced chemical equation arises when the number of atoms or ions of the reactant side equals the number of atoms or ions of the product side.

(e)

Expert Solution
Check Mark

Answer to Problem C13.4RE

The balanced chemical equation is:

  C3H8(g)  +  5O2(g)    3CO2(g)  +   4H2O(l)

Explanation of Solution

The unbalanced chemical reaction is as follows:

  C3H8(g)  +  O2(g)    CO2(g)  +   H2O(l)

To balance the above reaction, give coefficient 5 to O2, 3to CO2 and 4 to H2O thus, the balanced chemical equation will be:

  C3H8(g)  +  5O2(g)    3CO2(g)  +   4H2O(l)

(f)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given chemical reaction of hydrogen sulfide and oxygen gas needs to be determined

Concept Introduction:

The chemical reaction must be balanced such that it agrees with the law of mass conservation. A balanced chemical equation arises when the number of atoms or ions of the reactant side equals the number of atoms or ions of the product side.

(f)

Expert Solution
Check Mark

Answer to Problem C13.4RE

The balanced chemical equation is:

  2H2S(g)  +  3O2(g)    2SO2(g)  +   2H2O(l)

Explanation of Solution

The unbalanced chemical reaction is as follows:

  H2S(g)  +  O2(g)    SO2(g)  +   H2O(l)

To balance the above reaction, give coefficient 2 to H2S, 3to O2, 2 to SO2 and 2 to H2O thus, the balanced chemical equation will be:

  2H2S(g)  +  3O2(g)    2SO2(g)  +   2H2O(l)

(g)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given chemical reaction of hydrogen and oxygen gas needs to be determined.

Concept Introduction:

The chemical reaction must be balanced such that it agrees with the law of mass conservation. A balanced chemical equation arises when the number of atoms or ions of the reactant side equals the number of atoms or ions of the product side.

(g)

Expert Solution
Check Mark

Answer to Problem C13.4RE

The balanced chemical equation is:

  2H2(g)  +  O2(g)      2H2O(l)

Explanation of Solution

The unbalanced chemical reaction is as follows:

  H2(g)  +  O2(g)      H2O(l)

To balance the above reaction, give coefficient 2 to H2and 2 to H2O thus, the balanced chemical equation will be:

  2H2(g)  +  O2(g)      2H2O(l)

Chapter U4 Solutions

Living by Chemistry

Ch. U4.69 - Prob. 3ECh. U4.69 - Prob. 4ECh. U4.69 - Prob. 5ECh. U4.70 - Prob. 1TAICh. U4.70 - Prob. 1ECh. U4.70 - Prob. 2ECh. U4.70 - Prob. 3ECh. U4.70 - Prob. 4ECh. U4.70 - Prob. 6ECh. U4.71 - Prob. 1TAICh. U4.71 - Prob. 1ECh. U4.71 - Prob. 2ECh. U4.71 - Prob. 3ECh. U4.71 - Prob. 4ECh. U4.71 - Prob. 5ECh. U4.71 - Prob. 6ECh. U4.71 - Prob. 7ECh. U4.72 - Prob. 1TAICh. U4.72 - Prob. 1ECh. U4.72 - Prob. 2ECh. U4.72 - Prob. 3ECh. U4.72 - Prob. 4ECh. U4.73 - Prob. 1TAICh. U4.73 - Prob. 1ECh. U4.73 - Prob. 2ECh. U4.73 - Prob. 3ECh. U4.73 - Prob. 5ECh. U4.73 - Prob. 6ECh. U4.73 - Prob. 7ECh. U4.74 - Prob. 1TAICh. U4.74 - Prob. 1ECh. U4.74 - Prob. 2ECh. U4.74 - Prob. 4ECh. U4.74 - Prob. 5ECh. U4.75 - Prob. 1TAICh. U4.75 - Prob. 1ECh. U4.75 - Prob. 2ECh. U4.75 - Prob. 4ECh. U4.75 - Prob. 5ECh. U4.75 - Prob. 6ECh. U4.75 - Prob. 7ECh. U4.75 - Prob. 8ECh. U4.75 - Prob. 9ECh. U4.75 - Prob. 10ECh. U4.75 - Prob. 11ECh. U4.76 - Prob. 1TAICh. U4.76 - Prob. 1ECh. U4.76 - Prob. 2ECh. U4.76 - Prob. 3ECh. U4.76 - Prob. 4ECh. U4.76 - Prob. 5ECh. U4.76 - Prob. 6ECh. U4.76 - Prob. 7ECh. U4.76 - Prob. 8ECh. U4.77 - Prob. 1TAICh. U4.77 - Prob. 1ECh. U4.77 - Prob. 2ECh. U4.77 - Prob. 3ECh. U4.77 - Prob. 4ECh. U4.77 - Prob. 5ECh. U4.77 - Prob. 6ECh. U4.77 - Prob. 7ECh. U4.77 - Prob. 8ECh. U4.78 - Prob. 1TAICh. U4.78 - Prob. 1ECh. U4.78 - Prob. 2ECh. U4.78 - Prob. 3ECh. U4.78 - Prob. 4ECh. U4.78 - Prob. 5ECh. U4.78 - Prob. 6ECh. U4.78 - Prob. 7ECh. U4.78 - Prob. 8ECh. U4.79 - Prob. 1TAICh. U4.79 - Prob. 1ECh. U4.79 - Prob. 2ECh. U4.79 - Prob. 3ECh. U4.79 - Prob. 4ECh. U4.80 - Prob. 1TAICh. U4.80 - Prob. 1ECh. U4.80 - Prob. 2ECh. U4.80 - Prob. 3ECh. U4.80 - Prob. 4ECh. U4.80 - Prob. 5ECh. U4.80 - Prob. 6ECh. U4.80 - Prob. 7ECh. U4.80 - Prob. 8ECh. U4.80 - Prob. 9ECh. U4.80 - Prob. 10ECh. U4.81 - Prob. 1TAICh. U4.81 - Prob. 1ECh. U4.81 - Prob. 2ECh. U4.81 - Prob. 3ECh. U4.81 - Prob. 4ECh. U4.81 - Prob. 5ECh. U4.81 - Prob. 6ECh. U4.81 - Prob. 7ECh. U4.81 - Prob. 8ECh. U4.81 - Prob. 9ECh. U4.82 - Prob. 1TAICh. U4.82 - Prob. 1ECh. U4.82 - Prob. 2ECh. U4.82 - Prob. 3ECh. U4.82 - Prob. 4ECh. U4.82 - Prob. 5ECh. U4.82 - Prob. 6ECh. U4.82 - Prob. 7ECh. U4.82 - Prob. 8ECh. U4.83 - Prob. 1TAICh. U4.83 - Prob. 1ECh. U4.83 - Prob. 2ECh. U4.83 - Prob. 3ECh. U4.83 - Prob. 4ECh. U4.83 - Prob. 5ECh. U4.83 - Prob. 6ECh. U4.83 - Prob. 7ECh. U4.84 - Prob. 1TAICh. U4.84 - Prob. 1ECh. U4.84 - Prob. 2ECh. U4.84 - Prob. 3ECh. U4.84 - Prob. 5ECh. U4.84 - Prob. 6ECh. U4.84 - Prob. 7ECh. U4.85 - Prob. 1TAICh. U4.85 - Prob. 1ECh. U4.85 - Prob. 2ECh. U4.85 - Prob. 3ECh. U4.85 - Prob. 4ECh. U4.85 - Prob. 5ECh. U4.85 - Prob. 6ECh. U4.85 - Prob. 7ECh. U4.85 - Prob. 8ECh. U4.86 - Prob. 1TAICh. U4.86 - Prob. 1ECh. U4.86 - Prob. 2ECh. U4.86 - Prob. 3ECh. U4.86 - Prob. 4ECh. U4.86 - Prob. 6ECh. U4.86 - Prob. 7ECh. U4.86 - Prob. 8ECh. U4.87 - Prob. 1TAICh. U4.87 - Prob. 1ECh. U4.87 - Prob. 2ECh. U4.87 - Prob. 3ECh. U4.87 - Prob. 4ECh. U4.87 - Prob. 5ECh. U4.87 - Prob. 6ECh. U4.87 - Prob. 7ECh. U4.87 - Prob. 8ECh. U4.88 - Prob. 1TAICh. U4.88 - Prob. 1ECh. U4.88 - Prob. 2ECh. U4.88 - Prob. 4ECh. U4.88 - Prob. 5ECh. U4.88 - Prob. 6ECh. U4.88 - Prob. 7ECh. U4.88 - Prob. 8ECh. U4.89 - Prob. 1TAICh. U4.89 - Prob. 1ECh. U4.89 - Prob. 2ECh. U4.89 - Prob. 3ECh. U4.89 - Prob. 4ECh. U4.89 - Prob. 5ECh. U4.89 - Prob. 6ECh. U4.90 - Prob. 1ECh. U4.90 - Prob. 2ECh. U4.90 - Prob. 3ECh. U4.90 - Prob. 4ECh. U4.90 - Prob. 5ECh. U4.90 - Prob. 6ECh. U4.90 - Prob. 7ECh. U4.91 - Prob. 1ECh. U4.91 - Prob. 2ECh. U4.91 - Prob. 3ECh. U4.91 - Prob. 5ECh. U4.91 - Prob. 6ECh. U4.92 - Prob. 1TAICh. U4.92 - Prob. 1ECh. U4.92 - Prob. 2ECh. U4.92 - Prob. 3ECh. U4.92 - Prob. 4ECh. U4.93 - Prob. 1TAICh. U4.93 - Prob. 1ECh. U4.93 - Prob. 2ECh. U4.93 - Prob. 4ECh. U4.93 - Prob. 5ECh. U4.93 - Prob. 6ECh. U4 - Prob. C13.3RECh. U4 - Prob. C13.4RECh. U4 - Prob. C14.1RECh. U4 - Prob. C14.2RECh. U4 - Prob. C14.3RECh. U4 - Prob. C14.5RECh. U4 - Prob. C14.6RECh. U4 - Prob. C15.1RECh. U4 - Prob. C15.2RECh. U4 - Prob. C15.3RECh. U4 - Prob. C15.4RECh. U4 - Prob. C15.5RECh. U4 - Prob. C15.6RECh. U4 - Prob. C15.7RECh. U4 - Prob. C15.8RECh. U4 - Prob. C16.1RECh. U4 - Prob. C16.2RECh. U4 - Prob. C16.3RECh. U4 - Prob. C16.4RECh. U4 - Prob. C17.1RECh. U4 - Prob. C17.2RECh. U4 - Prob. C17.3RECh. U4 - Prob. 1RECh. U4 - Prob. 4RECh. U4 - Prob. 5RECh. U4 - Prob. 6RECh. U4 - Prob. 7RECh. U4 - Prob. 8RECh. U4 - Prob. 9RECh. U4 - Prob. 10RECh. U4 - Prob. 11RECh. U4 - Prob. 12RECh. U4 - Prob. 1STPCh. U4 - Prob. 2STPCh. U4 - Prob. 3STPCh. U4 - Prob. 4STPCh. U4 - Prob. 5STPCh. U4 - Prob. 6STPCh. U4 - Prob. 7STPCh. U4 - Prob. 8STPCh. U4 - Prob. 9STPCh. U4 - Prob. 10STPCh. U4 - Prob. 11STPCh. U4 - Prob. 12STPCh. U4 - Prob. 13STPCh. U4 - Prob. 14STPCh. U4 - Prob. 15STPCh. U4 - Prob. 16STPCh. U4 - Prob. 17STPCh. U4 - Prob. 18STPCh. U4 - Prob. 19STPCh. U4 - Prob. 20STP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Types of Matter: Elements, Compounds and Mixtures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=dggHWvFJ8Xs;License: Standard YouTube License, CC-BY