Answer:- Formula of the hydrate is 
and it's name is Iron(III)sulfate pentahydrate.
Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.
Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.
We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:
= 
= 
Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.
= 1
= 5
There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is and the name of the hydrate is Iron(III)sulfate pentahydrate.
Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol
Write an balance the equation
Na2O + H2O -> 2 NaOH
Calculate the molecular mass of Na2O and NaOH from the atomic mass from the periodic table.
Na = 23
O=16
H=1
Na2O = 23 * 2 + 16 = 62
NaOH = 23+16+1= 40
For the stoichiometry of the reaction one mole of Na2O = 62g produce two mol of NaOH = 2* 40= 80 g
120 g Na2O x 80g NaOH / 62g Na2O=
154.8 g NaOH
Answer:
3 Pb(NO3)2 + Al2(SO4)3 = 2 Al(NO3)3 + 3 PbSO4
Explanation:
I think your equation is incorrect? This is balanced and the sum of coefficients is 9.
Answer:
i don't think you can, but if you do know let me know
