Answer:
Ni (s) is oxidized because it loses electrons. This makes it the reducing agent.
CuCl₂ (aq) is reduced because it gains electrons. This makes it the oxidizing agent.
Explanation:
In the reaction:
Ni(s) + CuCl₂(aq) → Cu(s) + NiCl₂(aq)
Ni (s) is oxidized because it loses electrons. This makes it the reducing agent.
------------> Ni (s) has an oxidation number of 0. All lone atoms (that are not ions) have a 0 oxidation number.
-------------> In NiCl₂ (aq), chlorine must have an oxidation number of -1, and because there are two chorine atoms, they contribute a total -2 charge. To make the overall charge neutral, nickel must have an oxidation number of +2 to balance the charges.
--------------> Because nickel goes from an oxidation number of 0 to +2, this indicates that it lost electrons and was reduced.
CuCl₂ (aq) is reduced because it gains electrons. This makes it the oxidizing agent.
-------------> In CuCl₂ (aq), chlorine must have an oxidation number of -1, and because there are two chorine atoms, they contribute a total -2 charge. To make the overall charge neutral, copper must have an oxidation number of +2 to balance the charges.
------------> Cu (s) has an oxidation number of 0. All lone atoms (that are not ions) have a 0 oxidation number.
--------------> Because copper goes from an oxidation number of +2 to 0, this indicates that it gained electrons and was oxidized.
Avegadro's number = 6.02 x10^23 atoms
so
3.5g x 1mol/63.55g Cu x 6.02 x 10^23/ 1mol=3.32 x 10^22 atoms
Answer:
A scientific claim is a statement about the results of your experiment that is supported by evidence gathered during the experiment and reasoning that explains how the evidence is connected to the claim. For example: Claim: Air is matter.
Explanation:
Answer:
Explanation:
One way to calculate the lattice energy is to use Hess's Law.
The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:
Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?
We must generate this reaction rom the equations given.
(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹
(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹
(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹
(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹
(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹
Now, we put these equations together to get the lattice energy.
<u>E/kJ </u>
(5) Li⁺(g) +e⁻ ⟶ Li(g) 520
(6) Li(g) ⟶ Li(s) -161
(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409
(8) Cl(g) ⟶ ½Cl₂(g) -121.5
(9) Cl⁻(g) ⟶ Cl(g) + e⁻ <u>+349</u>
Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862
The lattice energy of LiCl is 
.
Gases in which the molecules that make it up naturally consist of two atoms of the same type.
