3 ed ugen uden mainfr.un Reade Wiley PLUS Abstract l Digital Library Weebly ebsi
ID: 965808 • Letter: 3
Question
3 ed ugen uden mainfr.un Reade Wiley PLUS Abstract l Digital Library Weebly ebsite Creation Made Easy Chegg Study I Guided S d Stud WileyPLUS Wiley PLUS: My Wiley PLUS I Help I Contact Us I Log Out Wiley PLUS Jespersen, The Molecular Nature of Matter, 7e GENERAL CHEMISTRY- SCIENCE MAJORS (CHE 1101/1102) Home Read, Study & Practice Gradebook ORION Assignment Open Assignment FULL SCREEN PRINTER ON BACK NEXT ASSIGNMENT RESOURCES Review Problem 16.094 Chapter 16 How many moles of NH3 must be dissolved in water to give 541.0 mL of solution with a pH of 11.10? Kb of ammonia is 1 M R 16.029 8 x 10 M R 16.030 M R 16.033 mol NH 6.036 16.050 the tolerance is +/-2% 16.052 16.060 16.064 Revi 16.066 By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. 16.069 M Review Problem 16.076 16.094 Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER 16.110 Earn Maximum Points available only if you Practi 6.09 answer this question correctly in four Practi 16.1 attempts or less. Practi 6.14 Practi 6.16 M T 16, 178 Copyright C 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved Review Score Review Results by Study Objective License Agreement l Privacy Policy I 2000-2016 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons Inc. 4.17.3.3Explanation / Answer
Kb = [NH4+][OH-]/[NH3]
pOH = 14-1.11 = 2.9
[OH-] = 10^-pOH = 10^-2.9 = 0.0012589
then
1.8*10^-5 = (0.0012589)(0.0012589)/(M-0.0012589)
M = 0.0012589^2 / (1.8*10^-5 ) + 0.0012589
M = 0.08930496
mol = MV = 0.08930496*0.541 = 0.0483139
mass = mol*mW = 0.0483139*17 = 0.8213363
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.