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FrozenT [24]
3 years ago
6

If the standard solutions had unknowingly been made up to be 0.0024 m agno3 and 0.0040 m k2cro4, would this have affected your r

esults? how?
Chemistry
2 answers:
Mnenie [13.5K]3 years ago
6 0
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with 
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
OlgaM077 [116]3 years ago
6 0

If the standard solutions had unknowingly been made up to be 0.0024 M AgNO₃ and 0.0040 M K₂CrO₄ the chemical equation would be:  

2AgNO₃ + K₂CrO₄ → Ag₂CrO₄ (s) + 2KNO₃

By mixing 0.0024 M AgNO₃ and 0.004 M K₂CrO₄ one would have Ag₂CrO₄ precipitated out.  

Thus, one is left with 0.0024 M KNO₃ mixed with (0.004 -0.0024 / 2) M = 0.0028 M of K₂CrO₄.  

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Given: 
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