1. In the laboratory, a student adds 44.6 mL of water to 10.8 mL of a 0.763 M pe
ID: 891234 • Letter: 1
Question
1. In the laboratory, a student adds 44.6 mL of water to 10.8 mL of a 0.763 M perchloric acid solution. What is the concentration of the diluted solution ? ______________ M
2. In the laboratory a student combines 24.3 mL of a 0.390 M copper(II) sulfate solution with 22.3 mL of a 0.302 Mpotassium sulfate solution.
What is the final concentration of sulfate anion ? ______________ M
3. Determine the oxidation state for each of the elements below.
The oxidation state of
sulfur
in
sulfite ion
SO32-
is _____
.
The oxidation state of
sulfur
in
sulfur solid
S8
is ________
.
The oxidation state of
carbon
in
oxalic acid
H2C2O4
is
______ .
The oxidation state of
...sulfur
...in
...sulfite ion
SO32-
is _____
....
The oxidation state of
sulfur
in
sulfur solid
S8
is ________
.
The oxidation state of
carbon
in
oxalic acid
H2C2O4
is
______ .
Explanation / Answer
1) 44.6 mL of water to 10.8 mL of a 0.763 M perchloric acid solution. What is the concentration of the diluted solution ?
perchloric acid H2o
M1=0.763 V1= 10.8 M2= ? V2= 44.6 ml
M1V1=M2V2
M2=[M1V1 ]/V2
M2=[0.763X10.8 ] / [44.6]
= 0.185 M
the concentration of the diluted solution is 0.185 M
2).
Cuso4----------------> cu+2 + So-24
n= [24.3x0.390]/1000 9.477x10-3 9.477x10-3
K2SO4 -----------------> 2K+ + SO-24
n= [22.3 X0.302] / 1000 6.7346x10-3 6.7346x10-3
Final concentration of sulfate ion = [9.477x10-3 x24.3] +[ 6.7346x10-3x22.3] / [24.3+22.3]
=[230.29x10-3] +[ 150.18x10-3] / [46.6]
= [380.47x10-3] /46.6
Final concentration of sulfate ion = 8.1645x10-3
3)
The oxidation state of Sulphur in SO-23
X+3(-2) = -2
X-6=-2
X=6-2
The oxidation state of Sulphur in SO-23 is +4
The oxidation state of Sulphur in S8 Zero
The oxidation state of carbon in H2C2O4
=2(1) + 2x +4(-2) =0
2x=6
x=+3
The oxidation state of carbon in H2C2O4 is +3
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