What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?
Cr3+ and Cr2O2−7 are formed
CrO2−4 is reduced to +3 state of Cr
Cr2O2−7 and H2O are formed
CrO2−4 is oxidized to +7 state of Cr
A
Cr2O2−7 and H2O are formed
B
CrO2−4 is oxidized to +7 state of Cr
C
Cr3+ and Cr2O2−7 are formed
D
CrO2−4 is reduced to +3 state of Cr
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Solution
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A solution of potassium chromate, when treated with an excess of dilute nitric acid, Cr2O2−7 and H2O are formed.
2K2CrO4+2HNO3→K2Cr2O7+2KNO3+H2O
Hence, Cr2O2−7 and H2O are formed.
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