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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 44p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 44p

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# Write a balanced chemical equation showing how each metal oxide reacts with HCl.(a)

ISBN: 9780321910295 34

## Solution for problem 44P Chapter 14

Introductory Chemistry | 5th Edition

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Problem 44P

Write a balanced chemical equation showing how each metal oxide reacts with HCl.

(a) SrO

(b) $$\mathrm {Na_2O}$$

(c) $$\mathrm {Li_2O}$$

(d) BaO

Equation Transcription:

Text Transcription:

Na_2O

Li_2O

Step-by-Step Solution:
Step 1 of 3

Solution 44P:

Here, we are going to write a balanced chemical equation.

(a) SrO

The balanced chemical equation of SrO with HCl.

SrO(s) + 2HCl(aq) ---------->SrCl2(s) +H2O(l)

(b) Na2O

The balanced chemical equation of Na2O with HCl.

Na2O(s) + 2HCl(aq) ---------->2NaCl(s) +H2O(l)

(c) Li2O

The balanced chemical equation of Li2O with HCl.

Li2O(s) + 2HCl(aq) ---------->2LiCl(s) +H2O(l)

(d) BaO

The balanced chemical equation of BaO with HCl.

BaO(s) + 2HCl(aq) ---------->BaCl2(s) +H2O(l)

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. This full solution covers the following key subjects: balanced, bao, chemical, equation, HCL. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. Since the solution to 44P from 14 chapter was answered, more than 1857 students have viewed the full step-by-step answer. The answer to “?Write a balanced chemical equation showing how each metal oxide reacts with HCl.(a) SrO(b) $$\mathrm {Na_2O}$$(c) $$\mathrm {Li_2O}$$(d) BaOEquation Transcription:Text Transcription:Na_2OLi_2O” is broken down into a number of easy to follow steps, and 21 words. The full step-by-step solution to problem: 44P from chapter: 14 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM.

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