First, we can neglect the Ka2 value and use Ka1:
according to this equation and by using the ICE table:
H2Se ⇄ H+ +HSe-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X
so,
Ka1 = [H+][HSe-] / [H2Se]
so by substitution:
1.3 x 10^-4 = X*X / (0.4 -X) by solving for X
∴X = 0.00715
∴[H+] = 0.00715 m
∴PH = -㏒[H+]
= -㏒ 0.00715
= 2.15
Yes, it is correct...........
First, we calculate the number of moles of each reactant using the formula:
Moles = mass / molecular weight
CaO:
Moles = 7.15/56 = 0.128
Water:
Moles = 152/18 = 8.44
The reaction equation shows that the reactants must be present in an equal number of moles, so CaO will be the limiting reactant and 0.128 mole of calcium hydroxide will form.
The energy released is given by:
Heat of reaction * number of moles
= -64.8 * 0.128
= -8.29 kJ
8.29 kJ of energy will be released
<u>Given:</u>
% Al = 35.94
% S = 64.06
<u>To determine:</u>
Empirical formula of a compound with the above composition
<u>Explanation:</u>
Atomic wt of Al = 27 g/mol
Atomic wt of S = 32 g/mol
Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g
# moles of Al = 35.94/27 = 1.331
# moles of S = 64.06/32 = 2.002
Divide by the smallest # moles:
Al = 1.331/1.331 = 1
S = 2.002/1,331 = 1.5 ≅ 2
Empirical formula = AlS₂
