- 1(current)
Q. No.If CrO42- (yellow) changes to Cr2O72- (orange) in pH =x and vice versa in pH=y, then x and y are:
1. 6,8
2. 6,5
3. 8,6
4. 7,7
1. \(IO^-_3\)
2. \(I_2\)
3. \(IO^-\)
4. \(IO^-_4\)
| 1. | 1, 3, 4, 5 | 2. | 3, 2, 1, 4 |
| 3. | 1, 5, 3, 7 | 4. | 4, 3, 2, 1 |
| 1. | +1 to +3 | 2. | +2 to +4 |
| 3. | +3 to +5 | 4. | +4 to +6 |
| 1. | \(\mathrm{C}_2 \mathrm{H}_6 \) | 2. | \(\mathrm{CH}_3 \mathrm{COOH} \) |
| 3. | \(\mathrm{K}^{+} \mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} \) | 4. | \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5\) |
In the titration of KMnO₄ or its use as an oxidizing agent, the medium is acidified with dilute H₂SO₄ rather than HCl because:
| 1. | Both HCl and KMnO4 act as oxidising agents |
| 2. | KMnO4 oxidises HCl into Cl2, which is also an oxidising agent |
| 3. | KMnO4 is a weaker oxidising agent than HCl |
| 4. | KMnO4 acts as a reducing agent in the presence of HCl |
1. Green
2. Black
3. Blue
4. Red
The fusion of chromite ore (FeCr2O4) with Na2CO3 in air gives a yellow solution upon the addition of water. Subsequent treatment with H2SO4 produces an orange solution.
The yellow and orange colours, respectively, are due to the formation of:
1. Na2CrO4 and Na2Cr2O7
2. Cr(OH)3 and Na2Cr2O7
3. Cr2(CO3)3 and Fe2(SO4)3
4. Cr(OH)3 and Na2CrO4
Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7solution?
1. The solution is decolorized.
2. SO2 is reduced.
3. Green coloured Cr2(SO4)3 is formed.
4. The solution turns blue.
- 1(current)
Q. No.
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