Answer:
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
2Fe + 3Cl2 → 2FeCl3
Explanation:
1. (SO4) 3 you see this 3 it means that 3 must be behind H2SO4. So now it's 3H2SO4.
2. If 3 is now behind one H2, it must be behind the other.
So now it's 3H2.
3. Al2 (SO4) 3 has 2 ahead of Al which means there will be 2Al in the reactants.
1. FeCl3 has 3 ahead of Cl, and Cl2 has 2. Which means that behind FeCl3 goes 2, and behind Cl2 goes 3 so now we have equated all Cl.
2. Since it is now 2FeCl3, we know that there must be 2 in the second Fe. It's 2Fe now.
Answer:
3.96 × 10⁻³ mol Pb
Explanation:
Step 1: Find molar mass
Pb (Lead) - 207.2 g/mol
Step 2: Use Dimensional Analysis
= 0.003962 mol Pb
Step 3: Simplify
We have 3 sig figs
0.003962 mol Pb ≈ 0.00396 mol Pb
0.00396 mol Pb = 3.96 × 10⁻³ mol Pb
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + <u>2H⁺ + 2e⁻</u>; E°´ = -0.031 V
<u>NAD⁺ + </u><u>2H⁺ + 2e⁻</u><u> ⇌ NADH + H⁺; </u> E°´ = <u> -0.320 V</u>
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
<u>FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; </u> E°´ = <u>-0.219 V
</u>
Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.
