Assignment Open Assignment ASSIGNMENT RESOURCES Additional Exercise 14.74 Chap 1

Question

Assignment Open Assignment ASSIGNMENT RESOURCES Additional Exercise 14.74 Chap 14 (a) How much concentrated sulfuric acid (H2so4) (17.8 M) is needed to prepare 8.4 L of 1.5 M sulfuric acid solution? Review Question 14.14 Review Question 14.17 L of 17.8 M H2SO4 must be diluted to a volume of 8.4 L (assume volumes are additive) Review Question 14.28 El Review Question 14.32 (b) How many moles of H2SO4 are in each ter of the original concentrate? Paired Exercise 145 Paired Exercise 14.10 mol H2SO4 in each mL of the original concentrate Paired Exercise 14.14 Paired Exercise 14.16. (c) How many moles are in uuut. ter of the diluted so n? Paired Exercise 14.18 mol H2SO4 in each mL of the diluted solution Paired Exercise 14.22 Paired Exercise 14.27 Paired Exercise 14.30 Paired Exercise 14.41 By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. Additional Exercise 14.45 Additional Exercise Question Attempts: Unlimited Additional Exercise 14.47 Additional Exercise Copyright C 2000-2017 by John Wiley & Sons, Inc. or related companies. All rights reserved. 14.56 dditional Exerci 14.74 Review Score Review Results by Study Objective License Agreement Priva Policy L o 2000-2017 ohn Wiley & Sons Inc All Rights Reserved. A Division of John Wiley & Sons, Inc MacBook Pro

Explanation / Answer

Given

1) Molarity M1 = 17.8 M

Volume V1 = ?

Molarity M2 = 1.5 M

Volume V2 = 8.4 L

M1 * V1 = M2 * V2

17.8 M * V1 = 1.5 M * 8.4 L

V1 = 0.708 L Answer

2) orginal concentrate is 17.8 mol/L

17.8 moles / 1000 ml = 0.0178 moles/ml Answer

3) orginal concentrate is 1.5 mol/L

1.5 mole/ 1000 ml = 0.0015 moles /ml Answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.