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analgon analgon
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12 years ago
35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO3)2. What
is the concentration of nitrate ion in the final solution?
A. 0.481 M
B. 0.296 M
C. 0.854 M
D. 1.10 M
E. 0.0295 M
help is deep appreciated
I know known the answer is A but don't known to get this answer
.35*0.22=.00892
.45*0.328=.01476
what do I do next
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#1
Valued Member
12 years ago
Quote from: analgon (12 years ago)
35.0 mL of 0.255 M nitric acid is added to 45.0 mL of 0.328 M Mg(NO3)2. What
is the concentration of nitrate ion in the final solution?
A. 0.481 M
B. 0.296 M
C. 0.854 M
D. 1.10 M
E. 0.0295 M
help is deep appreciated
I know known the answer is A but don't known to get this answer
.35*0.22=.00892
.45*0.328=.01476
what do I do next


Multiply .255 by .0350 L to get .00893 moles HNO3

Multiply .328 by .0450 L to get .0148 moles Mg(NO3)2
.
 but there are  .0296 moles of nitrate ions.

.00893 mol + .0296 mol = .0385 moles
.0350 L + .0450 L = .0800 L

.0385 mol / .0800 L = .481 M


so the option is A
I don't feel like riding until everything blurs.

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analgon Author
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#2
12 years ago
thank you
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