A beaker contains 10.888 mL of a 0.111 M solution of CSOH. This is titrated with 13.81 mL of a 0.0271 M solution of HN03. What is the final pH? (the answer should be entered with 3 significant figures; do not enter units; give answer in normal notation-examples include 1.23 and 120. ar -123 and 123. and 12.3) A Moving to another question will save this response. Question 5 o

A beaker contains 10.888 mL of a 0.111 M solution of CSOH. This is titrated with 13.81 mL of a 0.0271 M solution of HN03. What is the final pH? (the answer should be entered with 3 significant figures; do not enter units; give answer in normal notation-examples include 1.23 and 120. ar -123 and 123. and 12.3) A Moving to another question will save this response. Question 5 o

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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PROVISIONAL LEWIS STRUCTURES T01/S01 For each provisional structure, identify if it is the correct Lewis structure (C) or if it needs to be modified according to procedure MA or MB to obtain the correct Lewis structure. MA. Form multiple bonds to the central atom using lone pair on the terminal atom(s) until the central atom has an octet (or septet). MB. Form multiple bonds to the central atom using lone pair on the terminal atom(s) until the central atom has a formal charge of zero (0) or until two double bonds are formed. A provisional structure has the valence electrons distributed in order as (1) a bond between each pair of atoms, (2) an octet for each ligand (terminal) atom or ligand group (CH3, NH2 or OH) and (3) any remaining electrons on the central atom.
[3] e a: b: c: d: e: f: g: 1o
Asnwer is 0.81
CH3 HQ. HQ
1LDA 2) 10. Br OH GMH,SO,/A NH2 11. NH2 12.
C ?? D
How did they get 3? Do you always have to divide?
b. 3.20 x 10-4 4.72 x 108
(c) 1. THF H2C CEN 7. 2. H30* N(CH312 (d) 1. THF N' 2. H3O* 2016 Cmn Laung
1) NaBH4 2) H30* aLi
f) 0 Baz OEt ? OEt