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
Answer:
Explanation:
2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.
Answer:
a) No. of moles of hydrogen needed = 5.4 mol
b) Grams of ammonia produced = 27.2 g
Explanation:
a)
No. of moles of nitrogen = 1.80 mol
1 mole of nitrogen reacts with 3 moles of hydrogen
1.80 moles of nitrogen will react with
= 1.80 × 3 = 5.4 moles of hydrogen
b)
No. of moles of hydrogen = 2.4 mol
It is given that nitrogen is present in sufficient amount.
3 moles of hydrogen produce 2 moles of 
2.4 moles of hydrogen will produce
= 
Molar mass of ammonia = 17 g/mol
Mass in gram = No. of moles × Molar mass
Mass of ammonia in g = 1.6 × 17
= 27.2 g
The second volume : 42.2 L
<h3>Further explanation</h3>
Given
51.7 L at 27 C and 90.9 KPa
Required
The second volume
Solution
STP = P₂=1 atm, T₂=273 K
T₁ = 27 + 273 = 300 K
P₁ = 90.9 kPa = 0,897 atm
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.897 x 51.7/300 = 1 x V₂/273
V₂= 42.20 L
