Answer:
6.22 × 10⁻⁵
Explanation:
Step 1: Write the dissociation reaction
HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺
Step 2: Calculate the concentration of H⁺
The pH of the solution is 2.78.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M
Step 3: Calculate the molar concentration of the benzoic acid
We will use the following expression.
Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution
Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M
Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid
We will use the following expression.
Ka = [H⁺]²/Ca
Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵
Gravity. The rock is being pushed down and you are counteracting that force with a crowbar pushing it upward.
Answer:
The only nonmetal in group 14 is carbon.
Explanation:
I'd say b, but i'm not 100 percent sure.<span />
Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample
