Answer:
There is a temperature of 111 K required.
Explanation:
Step 1: Data given
Volume of hydrogen gas = 6.34 L
Pressure = 1.52 atm at 298 K
The volume of the gas at a pressure of 0.720 atm is 4.98 L
Step 2:
(P1*V1)/T1 = (P2*V2)/T2
⇒ with P1 = the pressure of the hydrogen gas = 1.52 atm
⇒ with V1 = the volume of the hydrogen gas = 6.34 L
⇒ with T1 = the temperature of the hydrogen gas = 298 K
⇒ with P2 = the changed pressure of the gas = 0.720 atm
⇒ with V2 = the changed volume of the gas = 4.98 L
⇒ with T2= the temperature of the gas at 0.720 atm and 4.98L
(1.52*6.34)/298 = (0.720*4.98)/T2
0.0323 = 3.5856/T2
T2 = 110.88 K ≈111K
There is a temperature of 111 K required.
Answer:
mass = 0.2982 grams
Explanation:
Given that;
volume = 100.0 mL = 0.1 L
Molarity =0.040 M
Since Molarity =
(x) no of moles = 0.040 × 0.1
= 0.004 moles
Also; to calculate the mass of KCl should you weigh to make this solution;
we have;
no of moles =
molar mass of KCl = 74.55 g/mol
∴
0.004 moles =
mass = 0.004 moles * 74.55
mass = 0.2982 grams
Hence, the mass of KCl should you weigh to make this solution is 0.2982 grams
Answer:
answer 2 I believe bbcc x
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06