Balance the following chemical equations: a) K2Cr2O7 + K2SO3 + H2SO4 → K2SO4 + Cr2(SO4)3 + H2O b) K2CrO4 + H2SO4 → K2SO4 + K2Cr2O7 + H2O c) KCrO2 + Br2 + KOH → K2CrO4 + KBr + H2O d) KOH + H2CrO4 → K2CrO4 + H2O

Let's balance each chemical equation step by step.

1. a) K₂Cr₂O₇ + K₂SO₃ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

First, let's identify the elements that change oxidation states. Chromium (Cr) is reduced from +6 in K₂Cr₂O₇ to +3 in Cr₂(SO₄)₃, and sulfur (S) is oxidized from +4 in K₂SO₃ to +6 in K₂SO₄ and Cr₂(SO₄)₃.

Balance the chromium atoms: K₂Cr₂O₇ → Cr₂(SO₄)₃

The change in oxidation state for Cr is (6 - 3) = 3. Since there are 2 Cr atoms, the total change is 2 * 3 = 6.

Balance the sulfur atoms: K₂SO₃ → K₂SO₄

The change in oxidation state for S is (6 - 4) = 2.

Now, equalize the changes in oxidation states by multiplying the compounds:

Multiply K₂Cr₂O₇ and Cr₂(SO₄)₃ by 2.

Multiply K₂SO₃ by 3.

So we have: K₂Cr₂O₇ + 3K₂SO₃ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Now, we balance the equation by adding coefficients to balance the remaining elements:

1 K₂Cr₂O₇ + 3 K₂SO₃ + 4 H₂SO₄ → 4 K₂SO₄ + 1 Cr₂(SO₄)₃ + 4 H₂O

The balanced equation is: K₂Cr₂O₇ + 3K₂SO₃ + 4H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O

2. b) K₂CrO₄ + H₂SO₄ → K₂SO₄ + K₂Cr₂O₇ + H₂O

To balance this equation, we need to consider that two chromate ions (CrO₄^2-) combine to form dichromate (Cr₂O₇^2-) in an acidic environment.

The balanced equation is: 2K₂CrO₄ + H₂SO₄ → K₂SO₄ + K₂Cr₂O₇ + H₂O

3. c) KCrO₂ + Br₂ + KOH → K₂CrO₄ + KBr + H₂O

In this reaction, chromium (Cr) is oxidized from +3 in KCrO₂ to +6 in K₂CrO₄, and bromine (Br) is reduced from 0 in Br₂ to -1 in KBr.

The balanced equation is: 2KCrO₂ + 3Br₂ + 8KOH → 2K₂CrO₄ + 6KBr + 4H₂O

4. d) KOH + H₂CrO₄ → K₂CrO₄ + H₂O

This is a neutralization reaction where potassium hydroxide (KOH) reacts with chromic acid (H₂CrO₄) to form potassium chromate (K₂CrO₄) and water (H₂O).

The balanced equation is: 2KOH + H₂CrO₄ → K₂CrO₄ + 2H₂O