← → C>tab<app.101edu.coSTARTING AMOUNTesc19XQ2Aktiv ChemistryW3.5#m3Determine the number of molecules of Cr₂O3 that form when 1.34 x 10³ g of oxygencompletely reacts according to the following equation:ADD FACTORX +x( )g Cr2.52 x 1025 3.78 x 10²56.022 x 102³3$4g Cr₂O3rf%54 Cr(s) + 3 O₂(g) -4g 0₂Question 23 of 38t21.68 x 1025mol Cr₂O3A16y2 Cr₂O3(s)4.24 × 10³&ANSWER7h1.34 x 10³mol CrO7F8C32.00152.00* 008RESET5molecules Cr₂O3ناسا3952.00kmol O₂F10OF11рFeb 24F126:14deleteSSubmit+X:backspace

Given a reaction , 4 Cr (s) + 3 O2 (g) → 2 Cr2O3 (s) We have to calculate the number molecules of…

Answered: Determine the number of molecules of…

Determine the number of molecules of Cr₂O, that form when 1.34 x 10³ g of oxygen completely reacts according to the following equation: 4 Cr(s) + 3 O₂(g) - 2 Cr₂O₂ (s)

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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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

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>
tab
<
app.101edu.co
STARTING AMOUNT
esc
1
9
X
Q
2
Aktiv Chemistry
W
3.5
#m
3
Determine the number of molecules of Cr₂O3 that form when 1.34 x 10³ g of oxygen
completely reacts according to the following equation:
ADD FACTOR
X +
x( )
g Cr
2.52 x 1025 3.78 x 10²5
6.022 x 102³3
$
4
g Cr₂O3
r
f
%
5
4 Cr(s) + 3 O₂(g) -
4
g 0₂
Question 23 of 38
t
2
1.68 x 1025
mol Cr₂O3
A
1
6
y
2 Cr₂O3(s)
4.24 × 10³
&
ANSWER
7
h
1.34 x 10³
mol Cr
O
7
F8
C
32.00
152.00
* 00
8
RESET
5
molecules Cr₂O3
ناسا
3
9
52.00
k
mol O₂
F10
O
F11
р
Feb 24
F12
6:14
delete
S
Submit
+
X
:
backspace

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