Answer:
An oxidising agent oxidises something else. Oxidation is loss of electrons (OIL RIG). That means that an oxidising agent takes electrons from that other substance.
Explanation:
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
Because a solubility is calculated for saturated solutions we can write:
36.3 g NaCl ---100g water
x g NaCl ---- 300.0 g water
x=36.3*300/100=108.9 g NaCl
Here, we are required to determine the volume of the earth which is 1.08326 × 10¹² km³ in liters.
<em>The volume of the earth is approximately</em>,
, 1.08326 × 10²⁴ liters
By conversion factors;
- <em>1dm³ = 1liter</em>
- However; <em>1km = 10000dm = 10⁴ </em><em>dm</em>
- Therefore, 1km³ = (10⁴)³ dm³.
Consequently, 1km³ = 10¹²dm³ = 10¹²liters.
The conversion factor from 1km³ to liters is therefore, c.f = 10¹²liters/km³
Therefore, the volume of the earth which is approximately, 1.08326 × 10¹² km³ can be expressed in liters as;
<em>1.08326 × 10¹² km³ × 10¹²liters/km³ </em>
The volume of the earth is approximately,
1.08326 × 10²⁴ liters.
Read more:
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