, THR CC14 formed in the first step is used as the reactant used in the second step.if 5.00 mol of CH4 reacts, what is the total amount of HCI producded. assume that C12 an HR in the presentin excess
Answer:
Ni^2+ is most likely
Ti^3+ is very unlikely
Explanation:
The Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in very many cases, but the degree of this favorability varies with the electronic configuration. In other words, for d1 there is only a small gap between the octahedral and tetrahedral lines, whereas at d3 and d8 is a very big gap. However, for d0, d5 high spin and d10, there is no crystal field stabilization energy difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:
d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10. This explains the answer choices above.
Ti^3+ being a d1 specie is least likely to exist in octahedral shape while Ni2+ a d8 specie is more likely to exist in octahedral shape.
Answer:
d. N
Explanation:
Chemical equation:
Pb(NO₃)₂(aq) + K₂SO₄(aq) → PbSO₄(s) + KNO₃(aq)
Balanced Chemical equation:
Pb(NO₃)₂(aq) + K₂SO₄(aq) → PbSO₄(s) + 2KNO₃(aq)
Ionic equation:
Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq)
Net ionic equation:
Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)
The NO₃⁻(aq) and K⁺(aq)are spectator ions that's why these are not written in net ionic equation. The PbSO₄ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.