I think the effect of increasing temperature would be; the equilbrium will shift back wards. Increase in temperature favors backward reaction since the forward reaction is exothermic and the backward reaction is endothermic. Therefore, the equilibrium will shift back wards, and there will be more reactants (H2 and Cl2) compared to the products
Explanation:
As density is defined as the mass of a substance divided by its volume.
Mathematically, Density = 
It is given that mass is 50 g and density is 0.934 
.
Hence, calculate the volume of methyl acetate as follows.
Density =
0.934 = 
Volume = 
or, = 
(as 1 
= 1 mL)
Thus, we can conclude that the volume of methyl acetate the student should pour out is .
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
Explanation:
No of molecules=0.500×6.023×10²³=3.011×10²³ molecules
