The solution for this problem is:
Get into moles first. .0560 grams over 540.8 grams per mole = 1.04 x l0^-4 moles
Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq)
Ksp = (Sr++)^3(As04^-3)^2
(Sr++) = 3 X 1.04 x l0^-4= 3.11 x l0^-4
(As04^-3) = 2 x 1.04 x l0^-4= 2.07 x l0^-4
Ksp = (1.04 x l0^-4)^3 (2.07 x l0^-4)^2 which equals 4.82 x 10^-20
We are given with
136 g P4
excess oxygen
The complete combustion reaction is
P4 + 5O2 => 2P2O5
Converting the amount of P4 to moles
136/123.9 = 1.098 moles
Using stoichiometry
moles P2O5 = 1.098 x 2 = 2.195 moles P2O5
Explanation: C) the air temp. at the top is lower
Answer:
0.534
Explanation:
Mole fraction can be calculated using the formula:
Mole fraction = number of moles of solute ÷ number of moles of solvent and solute (solution).
In this question, solute is dimethyl ether while the solvent is methanol.
Mole (n) = mass (M) ÷ molar mass (MM)
Mole of solute (dimethyl ether) = 148.5 ÷ 46.07
= 3.22moles.
Mole of solvent (methanol) = 90 ÷ 32.04
= 2.81moles.
Total moles of solute and solvent = 3.22 + 2.81 = 6.03moles.
Mole fraction of dimethyl ether = number of moles of dimethyl ether ÷ number of moles of solution (dimethyl ether + methanol)
Mole fraction = 3.22/6.03
= 0.534
