Assuming that the contents of the chamber ar ideal gases. We can use the relation 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
P2 = (1)(450)/ 48
P2 = 9.375 atm
The water gets cold because the ice cube is endothermic. Which means it takes in heat. It takes the warmth from the water and melts the ice cube which the ice cube releases coldness which cools down the water.
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
I’m pretty sure it’s NH4, ammonium
The answer is A.
The molecules in the air located.......below.
