HF has a higher [OH-] than a solution of 1.0 M HCl.
-log(0.5 M HF) = 0.3 pH
apexvs answer 1.0 x 10-4
6.38*10^-30C*m
pH = -log[H+] = -log(0.0005) = 3.3
8.0*10^-10 m
-log(0.5 M HF) = 0.3 pH
apexvs answer 1.0 x 10-4
The answer is 7,5g.
6.38*10^-30C*m
An 0.010 M HF solution gives pH = 2.6, so 0.0022 molesare present in 0.22 L
pH = -log[H+] = -log(0.0005) = 3.3
8.0*10^-10 m
10-4
concentration = molarity molarity= moles/ volume (in liters) H=1g F= 19g 1gH+ 19gF= 20gHF 20gHF x (1 mol HF/ 20gHF) = 1mol HF 2.0x10^2ml= 200ml x (1L/1000mL)= .200L Concentration= 1 mol HF/ .200L Concentration= 5M
true
0.13 is the concentration of the acetic acid solution.
It is a linear equation in the single variable, m. It has the solution m = 0