Answer: E=∆H*n= -40.6kj
Explanation:
V(CO) =15L=0.015M³
P=11200Pa
T=85C=358.15K
PV=nRT
n=(112000×0.015)/(8.314×358.15)
n(Co)= 0.564mol
V(Co)= 18.5L = 0.0185m³
P=744torr=98191.84Pa
T= 75C = 388.15k
PV=nRT
n= (99191.84×0.0185)/(8.314×348.15)
n(H2) = 0.634mol
n(CH30H) =1/2n(H2)=1/2×0.634mol
=0.317mol
∆H =∆Hf{CH3OH}-∆Hf(Co)
∆H=-238.6-(-110.5)
∆H = 128.1kj
E=∆H×n=-40.6kj.
He answer to what was is ii
To calculate the concentration of the base based on the titration, the concept used is the equal of number of equivalence of the acid used to that of the base. From this,
Na x Va = Nb x Vb
For HBr and KOH, molarity is equal to normality. Substituting the known values,
(0.75 M) x (22.6 mL - 0 mL) = Nb x (37.5 mL - 0.5 mL)
Nb = 0.46 N
Mb = 0.46 M
Thus, the concentration of the base is approximately 0.46 M.
