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.
Answer:
335.35 grams
Explanation:
Convert grams of H2SO4 to moles and get 4.08 moles.
Using stoichiometry, the chemical equation tells us that for every mole of H2SO4, one mole of Na2SO4 is produced. This means that, theoretically, you should get 4.08 moles.
Convert moles to grams and get 335.34 g.
However, if you were to actually perform this reaction, you would get less than the theoretical value.
Hope this helps! <3
Potassium dichromate reacts with sulfuric acid to form chromic acid, H₂CrO₄ which is a very strong oxidizing agent. The secondary alcohol, (<em>R</em>)-2-butanol will be oxidized in the presence of chromic acid, but it can only be as oxidized as far as the ketone, which is the product shown, 2-butanone.
Sodium borohydride is a reducing agent that will reduce a ketone or aldehyde to an alcohol. When sodium borohydride reacts with 2-butanone, it reduces it to 2-butanol. However, the alcohol is no longer chiral as it was in the beginning since the sodium borohydride can add a hydride to either face of the carbonyl, which results in a racemic mixture of alcohols. This explains why the product has the same refractive index and boiling point as (<em /><em>R</em>)-2-butanol, however, the product formed would no longer be optically active.