Answer : The equilibrium concentration of
at
is,
.
Solution : Given,
Equilibrium constant, 
Initial concentration of
= 0.260 m
Let, the 'x' mol/L of
are formed and at same time 'x' mol/L of
are also formed.
The equilibrium reaction is,

Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
![K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_2H_3O_2%5E-%5D%7D%7B%5BHC_2H_3O_2%5D%7D)
Now put all the given values in this expression, we get

By rearranging the terms, we get the value of 'x'.

Therefore, the equilibrium concentration of
at
is,
.
PV = nRT —> P = nRT/V
n = 0.250 moles of gas
R = 0.08206 L atm / mol K
T = 18 + 273 = 291 K
V = 7.50 L
P = (0.250)(0.08206)(291)/(7.50) = 7.96 atm
Answer:
Major organic products are- (a) propan-1-ol and (b) 2-methylpropan-2-ol
Explanation:
methyl magnesium bromide gives nucleophilic addition reaction with carbonyl group. Because methyl magnesium bromide is a strong nucleophile and carbonyl group is a strong electrophilic center.
Propanal contains an aldehyde group and propanone contains a ketone group. hence they both give nucleophilic addition with methyl magnesium bromide.
Dilute acid is added to protonate the alkoxide produced during nucleophilic addition.
Reactions are shown below.
Mg 10.9g/24.31=0.018/0.018=1
Cl 31.8g/35.45=0.897/0.018=49.83=50
O 57.3g/16.00=3.581/0.018=198.94=199
empirical formula=MgCl50O199
I think it's right