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Lab chemisty
03/27/2011

Five unlabeled bottles

Set 1:

A) colorless B) colorless C) blue D) blue E) colorless

A: Ba(NO3)2 B: AgNO3 C: CuSO4 D) CuCl2 E) KCl

Description how to identify solution:

_ We have two blue solution which are CuSO4 and CuCl2 or C and D, according to chemical reaction experiment, C didn’t have any reaction with other solution like B and D beside A, so if we look at the solubility chart, SO2- have only one precipitation with Ba2-. So we come to conclusion that C is CuSO4 and A is Ba(NO3)2

_ So now that we know C is CuSO4, other blue solution must be CuCl2 which is D, so according to chemical reaction experiment, D didn’t have any reaction with other solution beside B, if we look at solubility chart CuCl3 only have one precipitation with AgNO3. So we come to conclusion that B is AgNO3

_ E have only reaction with B, so now we know that B is AgNO3, and according to solubility chart Ag+ have two precipitation reaction, one of them is with CuCl2( based on previous information), so other precipitation must be with Kcl. So we come to conclusion that E is Kcl

Set 2:

A) LiBr B) HCl C) KOH D) Na2CO3 E) ZnCl2

Description how to identify solution:

_ According to chemical reaction experiment, D have two reaction which are one Gas ( B with D) and one precipitation ( D with E ). And if we look at solubility chart, Na2CO3 also have two reaction which are one gas reaction: CO3- with H+ and precipitation : CO3- with ZnCl2. So we come to conclusion that D is Na2CO3, B is Hcl because it’s gas reaction with Na2CO3, and E is ZnCl2 because it’s precipitation with Na2CO3.

_ According to chemical reaction experiment, C have only one precipitation reaction with E which we it’s ZnCl2. If we look at solubility chart, OH- also have one precipitation with ZnCl2. So we come to conclusion that C is KOH

_ A didn’t have any...
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