Answer:
Option A; V = 2.92 L
Explanation:
If we assume a lot of things, like:
The gas is an ideal gas.
The temperature is constant.
The gas does not interchange mass with the environment.
Then we have the relation:
P*V = n*R*T = constant.
Where:
P = pressure
V = volume
n = number of moles
R = constant of the ideal gas
T = temperature.
We know that when P = 0.55 atm, the volume is 5.31 L
Then:
(0.55 atm)*(5.31 L) = constant
Now, when the gas is at standard pressure ( P = 1 atm)
We still have the relation:
P*V = constant = (0.55 atm)*(5.31 L)
(1 atm)*V = (0.55 atm)*(5.31 L)
Now we only need to solve this for V.
V = (0.55 atm/ 1 atm)*(5.31 L) = 2.92 L
V = 2.92 L
Then the correct option is A.
Answer:
I' really don't know I'm sorry
Complete balanced equation: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
Ionized equation (with spectator ions):
2H⁺ + 2NO₃⁻ + Ca²⁺ + 2OH⁻ → Ca²⁺ + 2NO₃⁻ + 2H₂O
By eliminating the ions that are the same of both sides of the equation (spectator ions):
2H⁺ + 2OH⁻ → 2H₂O [Net Ionic Equation]
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
