Volume of 0 .108 M H 2 SO 4 solution needed to neutralize 15 .0 μ L solution of 0 .010 M Ca ( OH ) 2 has to be calculated. Concept Introduction: The dilution formula is given as follows: n 1 M 1 V 1 = n 2 M 2 V 2 Here, M 1 denotes the molarity of the solution before dilution. V 1 denotes the volume of the solution before dilution. M 2 denotes the molarity of solution after dilution. V 2 denotes the volume of the solution after dilution. n 1 and n 2 are n factors of acid and base respectively.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Predict the organic products that form in the reaction below: H. H+ + OH H+ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Explanation Check Click and drag to start drawing a structure. G X C © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Access +

111 Carbonyl Chem Choosing reagants for a Wittig reaction What would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. × ©2025 McGraw Hill LLC. All Rights Reserved. Terms of Use

A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. . If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T X O O лет-ле HO OH HO OH This transformation can't be done in one step.

Answer 2

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Answer 6

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

Answer 7

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

Answer 8

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

Answer 13

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Answer 14

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Answer 20

Ch. 12 - Prob. 12.1P Ch. 12 - Prob. 12.2P Ch. 12 - Prob. 12.3P Ch. 12 - Prob. 12.4P Ch. 12 - Prob. 12.5P Ch. 12 - Prob. 12.6P Ch. 12 - Prob. 12.7P Ch. 12 - Prob. 12.8P Ch. 12 - Prob. 12.9P Ch. 12 - Prob. 12.10P

Concept explainers

Videos

Answer to Problem 12.31P

Explanation of Solution

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 12.31P

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

Concept explainers

Videos

Answer to Problem 12.31P

Explanation of Solution

Interpretation: Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 12.31P

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).