How many equivalents in are in 2.00 L of a 5.00 M solution of Na2SO4?

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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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How many equivalents in are in 2.00 L of a 5.00 M solution of Na2SO4?
First we need the equation for dissolving the salt, Na2SO4 → 2 Na*1 + SO42
An equivalent is a mole of charges, either + or -, but not both, they are not added
the conversion factor is 1 mol =
eq
First calculate the moles
mol
L)(-
-) = 1.00 x
mol
L
Then convert into equivalents
eq
(1.00 x
mol)(-
-)3=
eq
1 mol
а. 0.00
b. 1.00
С. 2.00
d. 3.00
е. 4.00
f. 5.00
g. 6.00
h. 8.00
i. 10.00
j. 2000
k. 102
I. 103
m. 106
n. 109
o. 1012
р. 102
q. 10:3
r. 10-6
S. 10-9
t. 10-12
u. 108
V. 10-1
w. 101
Х. 30.0
у. О.100
z. 0.200
аа. 20.0
bb. 40.0
СС. 1.33.
dd. 0.267
Transcribed Image Text:How many equivalents in are in 2.00 L of a 5.00 M solution of Na2SO4? First we need the equation for dissolving the salt, Na2SO4 → 2 Na*1 + SO42 An equivalent is a mole of charges, either + or -, but not both, they are not added the conversion factor is 1 mol = eq First calculate the moles mol L)(- -) = 1.00 x mol L Then convert into equivalents eq (1.00 x mol)(- -)3= eq 1 mol а. 0.00 b. 1.00 С. 2.00 d. 3.00 е. 4.00 f. 5.00 g. 6.00 h. 8.00 i. 10.00 j. 2000 k. 102 I. 103 m. 106 n. 109 o. 1012 р. 102 q. 10:3 r. 10-6 S. 10-9 t. 10-12 u. 108 V. 10-1 w. 101 Х. 30.0 у. О.100 z. 0.200 аа. 20.0 bb. 40.0 СС. 1.33. dd. 0.267
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