The H3O+ in a 0.050M solution of Ba(OH)2 is calculated as below
write the equation for the dissociation of Ba(OH)2
Ba(OH)2 = Ba^2+ +2OH^-
calculate the OH- concentration
by use of mole ratio between Ba(OH)2 to OH^- which is 1:2 the concentration of OH = 0.050 x2 = 0.1 M
by use of the formula ( H3O+)(OH-) = 1 x10 ^-14
by making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-
substitute for OH-
H3O+ = (1 x10^-14 )/0.1
= 1 x10^-3 M
V1/T1 = V2/T2
24.5 + 273 = 298.5K
46.9 + 273 = 319.9K
95.8/298 = V2/319.9
V2 = 102.66L
Answer:
m = 29.6 grams
Explanation:
Given that,
Number of moles = 0.540
The molar mass of manganese = 54.93 g/mol
We know that,
Number of moles = given mass/molar mass

So, the required mass of the Manganese is equal to 29.6 grams.