The compounds that could exists between the ions K^+ and S^2- are K2S and K2S2.
<h3>What is an ionic compound?</h3>
An ionic compound is a commpound that is composed of ions. Let us note that ionic compounds are always such that the sum of the charges on the ions is zero.
Hence, the compounds that could exists between the ions K^+ and S^2- are K2S and K2S2.
Learn more about ionic compounds:brainly.com/question/9167977
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<h3>
Answer:</h3>
147.05 g/mol
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
CuClO₃
<u>Step 2: Find MM</u>
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CuClO₃ - 63.55 + 35.45 + 3(16.00) = 147.05 g/mol
Answer:
SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.
Explanation:
Oxidation reaction:
3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻
Reduction reaction:
Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)
Now, adding the oxidation and the reduction reactions we get the full net reaction:
Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)
Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.
<u>Therefore, SO₃²⁻ is the reducing agent. </u>
And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.
<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>