Answer:
pH = 13.09
Explanation:
Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15
Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15
K = Ksp X Kf
= 3*2*10^-15 * 10^15
= 6
Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M
Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)
Initial: 0 0.3 0
Change: -2x +x
Equilibrium: 0.3 - 2x x
K = Zn(OH)₄²⁻/[OH⁻]²
6 = x/(0.3 - 2x)²
6 = x/(0.3 -2x)(0.3 -2x)
6(0.09 -1.2x + 4x²) = x
0.54 - 7.2x + 24x² = x
24x² - 8.2x + 0.54 = 0
Upon solving as quadratic equation, we obtain;
x = 0.089
Therefore,
Concentration of (OH⁻) = 0.3 - 2x
= 0.3 -(2*0.089)
= 0.122
pOH = -log[OH⁻]
= -log 0.122
= 0.91
pH = 14-0.91
= 13.09
1 atom Mg, 2 atoms O and 2 atoms H.
1 + 2 + 2 = 5, so correct answer is C
Answer:
57.48%
Explanation:
Calculate the mass of 1 mole of malachite:
MM Cu = 63.55
MM O = 16.00
MM H = 1.01
MM C = 12.01
A mole of malachite has:
2 moles of Cu
5 moles of O
2 moles of H
1 mole of C
MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)
MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01
MW Malachite = 221.13
Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1
Now divide the mass of Cu by the mass of Malachite
The answer is 3.
That is if an atom has 5 electrons in its outer shell, then it has 3 unpaired electrons.
As the outer shell have 1 s orbital too and that is fully filled and not available for bonding so it must have 3 unpaired electrons.
So the number of unpaired electrons in an atom that is having 5 electrons in its outer shell is 3.
