We calculate for the number of moles of water given its mass by dividing the given mass by the molar mass.
n water = (36.04 g) / (18 g/mol)
n water = 2 mols
From the given balanced equation, every 6 moles of water produced will require 7 moles of oxygen.
n oxygen = (2 mols H2O) x (7 moles O2 / 6 moles H2O)
n oxygen = 2.33 mols O2
Answer:
S°m,298K = 85.184 J/Kmol
Explanation:
∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol
∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol
∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol
∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol
⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)
⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol
⇒ S°m,298K = 85.184 J/Kmol
Answer:
12.78 kJ
Explanation:
The correct balanced reaction would be

Mass of methanol = 
Moles of methanol can be obtained by dividing the mass of methanol with its molar mass 

Enthalpy change for the number of moles is given by


The change in enthalpy is 12.78 kJ.
Answer:
by using ideal gas law
Explanation:
ideal gas law:
PV=nRT
where:
P is pressure measured in Pascal (pa)
V is volume measured in letters (L)
n is number of moles
R is ideal gas constant
T is temperature measured in Kelvin (K)
by applying the given:
P(initial) V(initial)=nRT(initial)
P(final) V(final)=nRT(final)
nR is constant in both equations since same gas
then,
P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)
then by crossing multiply both equations
V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)
P(initial)=P(final)= 1 atm = 101325 pa
V(initial)= 6 L
T(initial) = 28°c = 28+273 kelvin
T(final) = 39°c = 39+273 kelvin
by substitution
V(final) = 6.21926 L
The answer to this is A i think.