PH = -log([H+])
[H+] = 10^(-pH)
[H+] = 10^(-8.78) = 1.65*10^-9
[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.
[OH-] = Kw/[H+] = (1.0*10^-14)/(1.65*10^-9) = 6.06*10^-6
The concentration of OH- ions is 6.1*10^-6 M.
Answer:
1.09 grams
Explanation:
According to the following chemical equation:
HF + NaNO₃ -> HNO₃ + NaF
1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:
MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF
1 mol HF x 19.9 g/mol HF = 20 g
MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF
1 mol NaF x 42 g/mol NaF = 42 g
Thus, from 20 g of HF are produced 42 g of NaF ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:
20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF
Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.
Answer:
96.3%
Explanation:
Data obtained from the question include:
Theoretical yield = 54g
Actual yield = 52g
Percentage yield =..?
The percentage yield can be obtain as follow:
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 52/54 x 100
Percentage yield = 96.3%
Therefore, the percentage yield of the reaction is 96.3%
