<span>Oxidation or reaction with oxygen is a chemical property</span>
Answer:
53.491 g/mol
Explanation:
Create the chemical compound and find each individual element's molar mass. Lastly, add them up.
Answer:
1. BF3 This is a trigonal planar molecule; the electron density is drawn into a cloud that circles the Boron, this is made nonpolar by the geometrically equivalent structure of the surrounding electronegative Fluorines.
2. H2O The 2 lone pairs of e- of Oxygen makes the O partially negative, the H’s, partially positive. Polar.
3. NF3 Lone pair on Nitrogen overwhelmed by the 3 incredibly electronegative Fluorines. Polar
4. CH3Br The “Soft Ion” of Bromine is negative; it is electronegative. Polar.
5. SO2 the lone pairs of Oxygen, at approximately 119°-120° angles to one another will form a reasonance structure; there will be more lone pairs about the Oxygen than the Sulfur; the Sulfur will be partially positive compared to the oxygens. Polar.
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>
