Answer:
It usually leads to more confidence in the results
<span>By definition:
pH = pKa + log [acetate]/ [acetic acid]
so
5.02 = 4.74 + log [acetate] / 10 mmole
10mmole = 10/1000 = 0.01 mole
5.02 = 4.74 + log [acetate] / 0.01
5.02 - 4.74 = 0.28 = log [acetate] /0.01
10^0.28 = </span><span>1.90546</span> = [acetate] / 0.01 <span>
[acetate] = 0.019 mole
= 19 millimoles
</span>
Sorry if I'm wrong but I think that it is B.
To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
<span>P2 = P1V1/V2</span>
<span>
</span>
<span>The correct answer is the first option. Pressure would increase. This can be seen from the equation above where V2 is indirectly proportional to P2.</span>
(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???
when the reaction equation is:
C(s) + H2O(g) ↔ H2(g) + CO(g)
∴ Kc = [H2] [CO] / [H2O]
and we have Kc = 0.0393 (given missing in the question)
when the O2 is added so, the reaction will be:
2H2(g) + O2(g) → 2H2O(g)
that means that 0.15 mol H2 gives 0.15 mol of H2O
∴ by using ICE table:
[H2O] [H2] [CO]
initial 0.57 + 0.15 0 0.15
change -X +X +X
Equ (0.72-X) X (0.15+X)
by substitution:
0.0393 = X (0.15+X) / (0.72-X) by solving for X
∴ X = 0.098
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
= 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
= 0.72 - 0.098
= 0.622 M
