<h3>
Answer:</h3>
82.11%
<h3>
Explanation:</h3>
We are given;
- Theoretical mass of the product is 137.5 g
- Actual mass of the product is 112.9 g
We are supposed to calculate the percentage yield
- We need to know how percentage yield is calculated;
- To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.
Thus;
% yield = (Actual mass ÷ Experimental mass) × 100%
= (112.9 g ÷ 137.5 g) × 100%
= 82.11%
Therefore, the percentage yield of the product is 82.11 %
Metals present in municipal waste water may still be present in treated sewage sludge IN CONCENTRATIONS THAT MAY AFFECT THE PUBLIC HEALTH. Sewage sludge is an end product of municipal waste water treatment and it contains many of the pollutant that are removed from the waste water.
The answer would be A. Genotype
Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
<u />
<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
