Suppose 4.95 g of lead (II) nitrate is dissolved in 250. mL of a 0.50 M aqueous solution of ammonium sulfate.Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the lead (II) nitrate is dissolved in it.Be sure your answer has the correct number of significant digits.MExplanationCheck00x10XSⒸ2023 McGraw Hill LLC. All Rights Reserved. Terms of UsePrivacy Center Accessi3

First, we need to calculate the number of moles of Pb(NO3)2 in the solution. Provided the mass of…

Answered: Suppose 4.95 g of lead(II) nitrate is…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

Suppose 4.95 g of lead (II) nitrate is dissolved in 250. mL of a 0.50 M aqueous solution of ammonium sulfate.
Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the lead (II) nitrate is dissolved in it.
Be sure your answer has the correct number of significant digits.
M
Explanation
Check
0
0x10
X
S
Ⓒ2023 McGraw Hill LLC. All Rights Reserved. Terms of Use
Privacy Center Accessi
3

Expert Solution

Step 1: Calculating number of moles of lead nitrate

First, we need to calculate the number of moles of Pb(NO₃)₂ in the solution.

Provided the mass of Pb( NO₃)₂ is 4.95 g and the molar mass of Pb( NO₃)₂ is 331.2 g/mol

by using the following formula we can calculate the moles of lead nitrate;

$table row cell N u m b e r italic space o f italic space m o l e s italic space P b left parenthesis N O subscript italic 3 right parenthesis subscript 2 italic space end cell italic equals cell fraction numerator G i v e n italic space m a s s italic space o f space P b left parenthesis N O subscript italic 3 right parenthesis subscript 2 italic space over denominator M o l e c u l a r italic space m a s s italic space o f space P b left parenthesis N O subscript italic 3 right parenthesis subscript 2 end fraction end cell row cell N u m b e r space o f space m o l e s space P b left parenthesis N O subscript italic 3 right parenthesis subscript 2 end cell equals cell space fraction numerator 4.95 space g over denominator 331.2 space g. m o l to the power of negative 1 end exponent space end fraction end cell row blank equals cell 0.0149 space m o l e s end cell end table$

Step by step

Solved in 4 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

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