CHEM110 PSU Homework 4 Exercise Assignment Help

Anonymous
timer Asked: Feb 22nd, 2019
account_balance_wallet $30

Question Description

i want from you to solve the problem in the file.

CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.39_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.46_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.51_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.55_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.59_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.59_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.39_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.46_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.51_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.55_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.39_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.46_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.51_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.55_pm.png
CHEM110 PSU Homework 4 Exercise Assignment Help
screen_shot_2019_02_21_at_10.31.59_pm.png

Tutor Answer

Aljon2017
School: Cornell University

Hi, please see the following file for the answers. All final answers are highlighted in yellow

SOLUTION:
(a) mol O2 = 7.5 mol KClO3 × (3 mol O2 / 2 mol KClO3) = 11 mol O2
(b) mol O2 = 7.5 mol H2O2 × (1 mol O2 / 2 mol H2O2) = 3.8 mol O2
(c) mol O2 = 7.5 mol HgO × (1 mol O2 / 2 mol Hg) = 3.8 mol O2
(d) mol O2 = 7.5 mol NaNO3 × (1 mol O2 / 2 mol NaNO3) = 3.8 mol O2
(e) mol O2 = 7.5 mol KClO4 × (2 mol O2 / 1 mol KClO4) = 15 mol O2

SOLUTION:
The balanced chemical reaction is as follows:
C6H12 + H2 → C6H14
moles H2 = 453 mol C6H12 × (1 mol H2 / 1 mol C6H14) = 453 mol H2

SOLUTION:
(a)
2C2H3Cl + 5O2 → 4CO2 + 2H2O + 2HCl

(b)
MW of C2H3Cl = 62.498 g/mol
35 g C2H3Cl = (35 g / 62.498 g/mol) = 0.56002 mol C2H3Cl
mol O2 required = 0.56002 mol C2H3Cl × (5 mol O2 / 2 mol C2H3Cl)
mol O2 required = 1.400 mol O2

(c)
25 g C2H3Cl = (25 g / 62.498 g/mol) = 0.40001 mol C2H3Cl

mol CO2 formed = 0.40001 mol C2H3Cl × (4 mol CO2 / 2 mol C2H3Cl) = 0.800026 mol CO2
MW of CO2 = 44.01 g /mol
mass o...

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Review

Anonymous
Good stuff. Would use again.

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