Chem 161-2009 Homework 5th week

Chem 161-2009 Homework 5th week - Chem 161-2009 Homework...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 161-2009 Homework 5 th Week Chapter 4: Reading: 4.3-4.6. Homework: 65,69,75,79,83,91,104 Chapter 5: Reading: 5.1-5.6. Homework: 5,21,23,25,31,37,41,47,95 65. Indicate whether the first-named substance in each change undergoes an oxidation, a reduction, or neither. Explain your reasoning. (a) Blue CrCl 2 (aq) changes to green CrCl 3 (aq) when exposed to air. (b) Yellow K 2 CrO 4 (aq) changes to orange K 2 Cr 2 O 7 (aq) when acidified. (c) Dinitrogen pentoxide produces nitric acid when it reacts with water. (a) Cr 2+ (O.N. +2) → Cr 3+ (O.N. +3)= oxidation, because the oxidation number increases. (b) CrO 4 2- (Cr O.N. +6) → Cr 2 O 7 2- (Cr O.N. +6) = Neither because the oxidation number doesn’t change. (c) N 2 O 5 (N O.N. +5) → HNO 3 (N O.N. +5) = Neither because the oxidation number doesn’t change. 69. Identify the oxidizing and reducing agents in Problem 67. Problem 67: (a) HCl + O 2 → Cl 2 + H 2 O (b) NO + H 2 → NH 3 + H 2 O (c) CH 4 + NO → N 2 + CO 2 + H 2 O (d) Ag + H + + NO 3 - → Ag + + H 2 O + NO (e) IO 4 - + I - + H + → I 2 + H 2 O +1 -1 0 0 +1 -2 (a) H Cl + O 2 Cl 2 + H 2 O Cl - oxidation state increased from -1 to 0; therefore oxidized; therefore, HCl is the reducing agent. O oxidation state reduced from 0 to -2; therefore reduced; therefore, O 2 is the oxidizing agent. +2 -2 0 -3 +1 +1 -2 (b) N O + H 2 N H 3 + H 2 O N oxidation state reduced from +2 to -3; therefore, reduced; therefore, NO is the oxidizing agent. H oxidation state increased from 0 to +1; therefore, oxidized; therefore, H 2 is the reducing agent. -4 +1 +2 -2 0 +4 -2 +1 -2 (c) C H 4 + N O N 2 + C O 2 + H 2 O C oxidation state increased from -4 to +4; therefore, oxidized; therefore, CH 4 is the reducing agent. N oxidation state decreased from +2 to 0; therefore reduced; therefore, NO is the oxidizing agent. 0 +1 +5 -2 +1 +1 -2 +2 -2 (d) Ag + H + + N O 3 - Ag + + H 2 O + N O Ag oxidation state increased from 0 to +1; therefore oxidized; therefore Ag is the reducing agent. N oxidation state reduced from +5 to +2; therefore, reduced; therefore NO 3 - is the oxidizing agent. +7 -2 -1 +1 0 +1 -2 (e) I O 4 - + I - + H + I 2 + H 2 O I oxidation state decreased from +7 to 0; therefore reduced; therefore IO 4 - is the oxidizing agent. I- oxidation state increased from -1 to 0; therefore, oxidized; therefore, I - is the reducing agent. Chem 161-2009 homework 5 th week 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
75. How many milliliters of 0.0195 M HCl are required to titrate (a) 25.00 mL of 0.0365 M KOH(aq), (b) 10.00 mL of 0.0116 M Ca(OH) 2 (aq), (c) 20.00 mL of 0.0225 M NH 3 (aq)? (a) HCl + KOH → KCl + HOH 0.0195M 0.0365M ?mL 25.00 mL MOH → mol OH - → mol H + → L H + → mL H + mol = M x V 0.0365 mol OH - /L OH - x 0.025 L x (1 mol H + /1 mol OH - ) x (1/0.0195 mol/L H + ) x (1000 mL/L) = 46.79 mL HCl Note: In water, strong acids and strong bases totally dissociate into their respective ions. Hence, HCl totally dissociates into H + + Cl - . Since we are beginning with 0.0195M HCl, we form 0.0195M H
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

Page1 / 7

Chem 161-2009 Homework 5th week - Chem 161-2009 Homework...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online