The volume of water needed to dissolve 0.0597 grams of barium chromate is ______ L.Assume no volume change upon addition of the solid. The mass of calcium hydroxide that is dissolved in 175 mL of a saturated solution is _____ grams. The mass of lead phosphate that is dissolved in 225 mL of a saturated solution is ______ grams.(The attached photo is an example of how to solve a similar problem) 1.34x10 4 moles PbSO4Ksp1/21.34x10 4 M =S =%3DL solutionWe'll carry 3 significant figures and round to 2 at the end.STEP 3 Determine the moles of PbSO, to dissolve:1 mol PbS040.0700 g PbSO4x2.31×101 moles PbSO4303.3 g PbS04STEP 4 Determine the volume needed:1 L solutionV = 2.31x104 moles PbSO41.7 L solution needed1.34×10 4 mol PbSO4 xL.The volume of water needed to dissolve 0.0700 grams of lead sulfate is 183x10^-8Assume no volume change upon addition of the solid.IncorrectSTEP 1 Determine the number of moles of PBSO4(s) that can dissolve in ONE liter:Let 's' = number of moles per liter that dissolve.%3DPBSO4 (s)= Pb2+ (aq) + SO,²- (aq)InitialsomeChange+ SEquilibriumsome+ s+ sSTEP 2 Set up the Ksp expression and solve for s:Ksp =[Pb2+] [So,2] = (s)(s) = s? = 1.8×108 (from the table)1.34x104 moles PbSO4Previous.Next1/2

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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...

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The volume of water needed to dissolve 0.0597 grams of barium chromate is ______ L.

Assume no volume change upon addition of the solid.
The mass of calcium hydroxide that is dissolved in 175 mL of a saturated solution is _____ grams.
The mass of lead phosphate that is dissolved in 225 mL of a saturated solution is ______ grams.

(The attached photo is an example of how to solve a similar problem)

1.34x10 4 moles PbSO4
Ksp
1/2
1.34x10 4 M =
S =
%3D
L solution
We'll carry 3 significant figures and round to 2 at the end.
STEP 3 Determine the moles of PbSO, to dissolve:
1 mol PbS04
0.0700 g PbSO4x
2.31×101 moles PbSO4
303.3 g PbS04
STEP 4 Determine the volume needed:
1 L solution
V = 2.31x104 moles PbSO4
1.7 L solution needed
1.34×10 4 mol PbSO4

xL.
The volume of water needed to dissolve 0.0700 grams of lead sulfate is 183x10^-8
Assume no volume change upon addition of the solid.
Incorrect
STEP 1 Determine the number of moles of PBSO4(s) that can dissolve in ONE liter:
Let 's' = number of moles per liter that dissolve.
%3D
PBSO4 (s)
= Pb2+ (aq) + SO,²- (aq)
Initial
some
Change
+ S
Equilibrium
some
+ s
+ s
STEP 2 Set up the Ksp expression and solve for s:
Ksp =[Pb2+] [So,2] = (s)(s) = s? = 1.8×108 (from the table)
1.34x104 moles PbSO4
Previous.
Next
1/2

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