Work is measured in joules. hope this helps!
Answer:
4 pairs are needed for the bonds, leaving 1 lone pair. Each double bond uses 2 bond pairs and can be thought of as a single unit. There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement.
The balanced half-reaction for the product that forms at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻ → I₂ + 2e- and 2H₂O → O₂ + 4H⁺ + 4e⁻
<h3>What is Balanced Chemical Equation ?</h3>
The equation during which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation is called balanced chemical equation.
Now write the equation for FeI₂
At cathode:
Fe⁺² + 2e⁻ → Fe(s) Eo = - 0.44 V
2H₂O + 2e⁻ → H₂ + 2OH⁻ Eo = - 0.827 V
It is easy to decrease Fe⁺² ions than the water, the product which is formed at cathode is Iron.
At anode:
2I⁻ → I₂ + 2e- Eo = - 0.54 V
2H₂O → O₂ + 4H⁺ + 4e⁻ Eo = -1.23 V
O₂ gas formed at anode.
Thus from the above conclusion we can say that The balanced half-reaction for the product at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻ → I₂ + 2e- and 2H₂O → O₂ + 4H⁺ + 4e⁻.
Learn more about the Balanced chemical equation here: brainly.com/question/26694427
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Answer:
a) 7.0 moles of NH3
b) 61.2 g of NH3
c) 4.15 g of H2
d) 8.9 ×10^19 molecules
Explanation:
Equation of the reaction;
N2(g) + 3H2(g) ⇄NH3(g)
a)
If 3 moles of H2 yields 1 mole of NH3
21 moles of H2 will yield 21 × 1 /3 = 7.0 moles of NH3
b)
1 mole of N2 yields 17 g of NH3
3.6 moles of N2 yields 3.6 moles × 17 g of NH3 = 61.2 g of NH3
c)
If 6g of H2 produces 17 g of NH3
xg of H2 will produce 11.76 g of NH3
x= 6 × 11.76/17
x= 4.15 g of H2
d)
If 6g of hydrogen yields 6.02 × 10^23 molecules of NH3
8.86 × 10^-4g of H2 yields 8.86 × 10^-4g × 6.02 × 10^23 /6 = 8.9 ×10^19 molecules
