The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
1425 mmHg.
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 1.5 L
Initial pressure (P1) = 1 atm
Final volume (V2) = 0.8 L
Final pressure (P2) =?
Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:
P1V1 = P2V2
1 × 1.5 = P2 × 0.8
1.5 = P2 × 0.8
Divide both side by 0.8
P2 = 1.5/0.8
P2 = 1.875 atm
Finally, we shall convert 1.875 atm to mmHg.
This can be obtained as follow:
1 atm = 760 mmHg
Therefore,
1.875 atm = 1.875 × 760 = 1425 mmHg.
Therefore, the new pressure of the gas is 1425 mmHg.
On point? Do you have any options?
Answer:
4 cups are about 0.25 gallons
Explanation:
To convert a cup measurement to a gallon measurement, divide the volume by the conversion ratio. The volume in gallons is equal to the cups divided by 16.
