I’m doing Chem right now, and want to make sure I’m doing it right before I continue. Can you correct my answers if they’re wrong? V₁M₂9.) 150.0 ml of 2.0 M NaBr from 6.0 M Stock Solution(0.50 M)(100.00 mL)2.0M(2.0 M) (150.0 mL) = (6.0 M) (V₂)6.0 M6.0 MV₁M₁M₂10.) 100.00 mL of 0.50 M nickel (11) Sulfate from 2.0 MStock solutionV₂ = 25=ory PoliciesV₂ = 50(2.0 M)(V₂)2.0MM₂V₁11.) 1.0 L of 0.10 M Sodium acetate from 2.5M StockSolution=V₂= 0.04(1.0L) (0.10 M) = (2.5 M) (V₂)2.5 M2.5 M12.) 500.0 mL of 0.80 M H₂SO4 from 12.014 stock.Solution(0.80 M) (500.0 mL) (12.0 M) (V₂) V₂ = 33.312.0M12.0M

Answered: Image /qna-images/answer/8cfe1ec3-5b9d-4f35-a30f-75a7986fb9f5.jpg

Answered: 150.0 ml of 2.0 M NaBr from (2.0 M)…

150.0 ml of 2.0 M NaBr from (2.0 M) (150.0 mL) = (6.0 6.0 M V₁ M₁ 100.00 mL of 0.50 M nicke OCK Solution (0.50 M)(100.00 mL) = ( 2.0M V₁ 316 OCE

Introductory Chemistry For Today

8th Edition

ISBN:9781285644561

Author:Seager

Publisher:Seager

Chapter7: Solutions And Colloids

Section: Chapter Questions

Problem 7.107E

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Question

I’m doing Chem right now, and want to make sure I’m doing it right before I continue. Can you correct my answers if they’re wrong?

V₁
M₂
9.) 150.0 ml of 2.0 M NaBr from 6.0 M Stock Solution
(0.50 M)(100.00 mL)
2.0M
(2.0 M) (150.0 mL) = (6.0 M) (V₂)
6.0 M
6.0 M
V₁
M₁
M₂
10.) 100.00 mL of 0.50 M nickel (11) Sulfate from 2.0 M
Stock solution
V₂ = 25
=
ory Policies
V₂ = 50
(2.0 M)(V₂)
2.0M
M₂
V₁
11.) 1.0 L of 0.10 M Sodium acetate from 2.5M Stock
Solution
=
V₂
= 0.04
(1.0L) (0.10 M) = (2.5 M) (V₂)
2.5 M
2.5 M
12.) 500.0 mL of 0.80 M H₂SO4 from 12.014 stock.
Solution
(0.80 M) (500.0 mL) (12.0 M) (V₂) V₂ = 33.3
12.0M
12.0M

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