a) To calculate: The [ O H − ] of stomach acid with 4 × 10 − 2 M [ H 3 O + ] The [ O H − ] of stomach acid is 2.5 × 1 0 − 13 M

Question

Want to see the full answer?

Answer 1

Question

Chapter 14.5, Problem 14.41QAP

Interpretation Introduction

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

Interpretation Introduction

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

Interpretation Introduction

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

Interpretation Introduction

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

So, the first image is what I'm trying to understand regarding my approach. The second image illustrates my teacher's method, and the third image includes my notes on the concepts behind these types of problems.

HAND DRAW

Draw a mental model for calcium chloride mixed with sodium phosphate

Answer 2

Question

Chapter 14.5, Problem 14.41QAP

Interpretation Introduction

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

Interpretation Introduction

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

Interpretation Introduction

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

Interpretation Introduction

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 14.5, Problem 14.41QAP

Interpretation Introduction

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

Interpretation Introduction

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

Interpretation Introduction

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

Interpretation Introduction

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 14.5, Problem 14.41QAP

Interpretation Introduction

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

Interpretation Introduction

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

Interpretation Introduction

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

Interpretation Introduction

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 14.5, Problem 14.41QAP

Interpretation Introduction

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

Interpretation Introduction

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

Interpretation Introduction

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

Interpretation Introduction

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Answer 6

Chapter 14.5, Problem 14.41QAP

Interpretation Introduction

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

Answer 7

Chapter 14.5, Problem 14.41QAP

Interpretation Introduction

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

Answer 8

Interpretation Introduction

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

Answer 9

Interpretation Introduction

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

Answer 10

Interpretation Introduction

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

Answer 11

Interpretation Introduction

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

Answer 12

Interpretation Introduction

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Answer 13

Interpretation Introduction

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 17

Ch. 14.1 - Indicate whether each of the following statements...Ch. 14.1 - Prob. 14.2QAP Ch. 14.1 - Prob. 14.3QAP Ch. 14.1 - Name each of the following acids or bases: a....Ch. 14.1 - Write formulas for each of the following acids and...Ch. 14.1 - Write formulas for each of the following acids and...Ch. 14.2 - Identify the reactant that is a Bronsted-Lowry...Ch. 14.2 - Prob. 14.8QAP Ch. 14.2 - Prob. 14.9QAP Ch. 14.2 - Prob. 14.10QAP

a) To calculate: The [ O H − ] of stomach acid with 4 × 10 − 2 M [ H 3 O + ] The [ O H − ] of stomach acid is 2.5 × 1 0 − 13 M

Concept explainers

Videos

a) To calculate: The $[O H^{-}]$ of stomach acid with $4 \times 10^{- 2} M [H_{3} O^{+}]$

The $[O H^{-}]$ of stomach acid is $2.5 \times 1 0^{- 13} M$

b) To Calculate: The $[O H^{-}]$ of Urine with $5 \times 10^{- 6} M [H_{3} O^{+}]$

The $[O H^{-}]$ of urine is $2 \times 10^{- 9} M$

c) To Calculate: The $[O H^{-}]$ of orange juice with $2 \times 10^{- 4} M [H_{3} O^{+}]$

The $[O H^{-}]$ of orange juice is $5 \times 10^{- 11} M$

d) To Calculate: The $[O H^{-}]$ of bile juice with $7.9 \times 10^{- 9} M [H_{3} O^{+}]$

The $[O H^{-}]$ of bile juice is $0.000001265 M$

Want to see the full answer?

Chapter 14 Solutions

a) To calculate: The [ O H − ] of stomach acid with 4 × 10 − 2 M [ H 3 O + ] The [ O H − ] of stomach acid is 2.5 × 1 0 − 13 M

Concept explainers

Videos