34. We mix 735 mL of 0.443 M Mg(CIO,), with 702 mL of 0.278 M Al(OH),. What is the molarity of Mg²+ in the combined solution after the reaction. (Assume that the volumes of the solutions are simply added to result in the final volume.) A) 0.914 M B) 0.159 M maes: (735) L0.443) = .325605 (702)(0.278) = .195156 My Crio, ³₂ mob ALLOH)3 mole 0.409 M C) D) 0.638 M E) 0.014 M 2 3Mg(co₂)2 + 2Al(OH)3 D3Mg(OH)₂ + 2A12C1E 32 560 5-1085 Limit? 3 mol •195156 amol =.097578 "ALLOH ma clo

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Molarity calculation question
please tell me what im doing wrong. thank you.
34. We mix 735 mL of 0.443 M Mg(CIO,), with 702 mL of 0.278 M Al(OH),. What is the molarity of Mg²+
in the combined solution after the reaction. (Assume that the volumes of the solutions are simply
added to result in the final volume.)
males: (735)(0.443) =
A) 0.914 M
= .325605 My Crio, 5₂ mole
(702)(0.278) = .195156 AILOH)3 mole
B)
0.159 M
C) 0.409 M
D) 0.638 M
E) 0.014 M
3Mg(co₂)2 + 2Al(OH)3 +3 Mg(OH)₂ + 2/1/(103) 3
Limit?
32 560 5-1085
.195156
amol
3 mol
mg clo
097578
Аксон
Limiting
Ⓒan't consumed
ing
097578 3mol al
img
тае
2mol
Al Cat)
41
mgcolt)
5 my initially brought in
325605 Imol mo
my (110) Imolimy
sing
325 605-.146367 =
n
179238
M = 1²/1/20
(735)+(702) 2
702) 2
=12473
JJ XX
Imol
I mol
C
146367
mol mg
in product
325605
mol mg in
reaget
-179238
mol
Transcribed Image Text:Molarity calculation question please tell me what im doing wrong. thank you. 34. We mix 735 mL of 0.443 M Mg(CIO,), with 702 mL of 0.278 M Al(OH),. What is the molarity of Mg²+ in the combined solution after the reaction. (Assume that the volumes of the solutions are simply added to result in the final volume.) males: (735)(0.443) = A) 0.914 M = .325605 My Crio, 5₂ mole (702)(0.278) = .195156 AILOH)3 mole B) 0.159 M C) 0.409 M D) 0.638 M E) 0.014 M 3Mg(co₂)2 + 2Al(OH)3 +3 Mg(OH)₂ + 2/1/(103) 3 Limit? 32 560 5-1085 .195156 amol 3 mol mg clo 097578 Аксон Limiting Ⓒan't consumed ing 097578 3mol al img тае 2mol Al Cat) 41 mgcolt) 5 my initially brought in 325605 Imol mo my (110) Imolimy sing 325 605-.146367 = n 179238 M = 1²/1/20 (735)+(702) 2 702) 2 =12473 JJ XX Imol I mol C 146367 mol mg in product 325605 mol mg in reaget -179238 mol
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