Ok yeah a ha yeah wow it is 13.543 g/× compare that with e 13.5
When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
= 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
= 14.28 * 17 = 242.76 g
Answer:
V₂ = 15.00 atm
Explanation:
Given data:
Initial pressure = 5.00 atm
Initial volume = 3.00 L
Final pressure = 760 mmHg ( 760/760 = 1 atm)
Final volume = ?
Solution:
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = 5.00 atm × 3.00 L / 1 atm
V₂ = 15.00 atm
