Explanation:
The initial concentrations for a mixture :
Acetic acid at equilibrium = 0.15 M
Ethanol at equilibrium = 0.15 M
Ethyl acetate at equilibrium = 0.40 M
Water at equilibrium = 0.40 M

Initially:
0.15 M 0.15 M 0.40 M 0.40 M
At equilibrium
(0.15-x)M (0.15-x) M (0.40+x) M (0.40+x) M
The equilibrium constant is given by expression
![K_c=\frac{[CH_3CO_2C_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3CO_2C_2H_5%5D%5BH_2O%5D%7D%7B%5BCH_3COOH%5D%5BC_2H_5OH%5D%7D)

Solving for x:
x = 0.0333
The equilibrium concentrations for a mixture :
Acetic acid at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M
Ethanol at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M
Ethyl acetate at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M
Water at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M
For getting the result of this problem, the knowledge of periodic table is very important. From the periodic table we come to know that. The knowledge of atomic mass of magnesium is also required to solve the problem.
1 mole of magnesium = 24.3 gm
88.1 moles of magnesium = (24.3 * 88.1) gms
= 2140.83 gms
So 2140.83 grams are there in 88.1 moles of magnesium.
The shared electrons are closer to Cl due to the greater electronegativity of chlorine.
Can you translate to english?
Use the equation d=m/v
your mass or "m" is 78 g
your volume or "v" is 60mL
if you plug those values into the equation it will look like this:
d=78/60
d=1.3g/mL should be what you come up with