For the reactionso2(g) + no2(g) so3(g) + no(g), the equilibrium constant is 18.0 at 1,200ºc. if 2.0 moles of so2 and 2.0 moles of no2 are placed in a 20. l container, what concentration of so3 will be present at equilibrium?
a) 0.081 mol/l
b) 0.019 mol/l
c) 0.11 mol/l
d) 1.00 mol/l
e) 18 mol/l

Answer : The correct option is, (A) 0.081 mol/L

Explanation : Given,

Equilibrium constant = 18.0

Initial concentration of =

Initial concentration of =

The balanced equilibrium reaction is,

Initial conc.    0.1          0.1             0             0

At eqm.        (0.1-x)     (0.1-x)          x             x

The expression of equilibrium constant for the reaction will be:

Now put all the values in this expression, we get :

By solving the term 'x' by quadratic equation, we get two value of 'x'.

From this we conclude that, the value of x = 0.131 at equilibrium can not be more than the initial concentration. So, the value of 'x' which is equal to 0.131 M is not consider.

The concentration of at equilibrium = x = 0.081 M

Therefore, the concentration of at equilibrium is, 0.081 M

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

plz give brainliest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%