Copper(I) compounds in aqueous solutions are unstable and disproportionate:
Cu₂SO₄ = Cu + CuSO₄
Necessary for dissolution of non-aqueous solvent.
M(Cu₂SO₄)=223.16 g/mol
m(Cu₂SO₄)=14.1 g
v=0.250 L
n(Cu₂SO₄)=m(Cu₂SO₄)/M(Cu₂SO₄)
c=n(Cu₂SO₄)/v=m(Cu₂SO₄)/(vM(Cu₂SO₄))
c=14.1/(0.250*223.16)=0.253 mol/L
0.253 M
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
- If the abundance of the first isotope is 68.037%, then the abundance of the second isotope is 100%-68.037%.
Substituting into the atomic mass formula,
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
