<u>Answer:</u> The given amount of iron reacts with 9.0 moles of and produce 6.0 moles of
<u>Explanation:</u>
We are given:
Moles of iron = 12.0 moles
The chemical equation for the rusting of iron follows:
By Stoichiometry of the reaction:
4 moles of iron reacts with 3 moles of oxygen gas
So, 12.0 moles of iron will react with = of oxygen gas
- <u>For iron (III) oxide:</u>
By Stoichiometry of the reaction:
4 moles of iron produces 2 moles of iron (III) oxide
So, 12.0 moles of iron will produce = of iron (III) oxide
Hence, the given amount of iron reacts with 9.0 moles of and produce 6.0 moles of
Answer:
pH =6.70
Explanation:
When Kw = 1x10^-14
[H3O+] = [OH-] --> 1x10-7
Kw =3.98x^10-14
3.98x^10-14 = [H3O+] . [OH-]
√ (3.98x^10-14) = [H3O+] = [OH-]
[H3O+] = 1.99×10^-7
- log [H3O+] = pH ---> -log 1.99x10^-7 = 6.70
The bond order of Mg2 molecule is given by:
Mg2 = 4e⁻= 3σs(2e⁻) 3σs*(2e⁻) 3σp(0) 3πp(0e⁻) 3πp*(0e⁻)
3σp*(0e⁻)
Bond Order = ½[Σ (bonding e-) - Σ (antibonding e-)]
bo = ½[ {σs(2e⁻)} - {σs*(2e⁻)}] = 0
the bond between Mg2 molecules are ionic bonding and it
stable.