Chemical reaction:
2H2S (g) ⇄ 2H2(g) + S2(g)
The equilibrium constant is given by: Kc = [H2]^2 * [S2] / [H2S]^2 = 1.67 * 10^ -7
The initial concentration is 0.0125 mol / 0.500 L = 0.0250 M
Make a table showing the initial concentrations, the change and the final concentrations of each species
2H2S (g) ⇄ 2H2(g) + S2(g)
start 0.0250M 0 0
change - 2x +2x + x
end 0.0250 - 2x 2x x
Kc = (2x)^2 (x) / (0.0250 - 2x)^2
Kc = 4x^3 / (0.0250 - 2x)^2
To solve that equation in an easy way you can assume that 2x is << 0.0250, which leads to
Kc = 4x^3 / (0.0250)^2 = 1.67 * 10^ -7
=> x^3 = 1.67 * 10^ -7 * 0.0250 / 4 = 2.6 * 10 ^-11
=> x = 2.97 * 10^ -4 M
With this you can check that your assumption that x << 0.0250 is good and continue.
From the table you know that the concentrations at equilibrium are:
[H2] = 2x = 2 * 2.97 * 10 ^ -4 M = 5.94 * 10 ^ -4
[S2] = x = 2.97 * 10^ -4 M
Answer:
2,2,1 being the coefficients
Explanation:
These type of question rely on the person answering the question knowing the formula of hydrogen peroxide. Hydrogen peroxide has a formula that is very similar to that of water except it has a two after the O, which makes it very easy to remember.
We can use combined gas law,
PV/T = k (constant)
Where, P is the pressure of the gas, V is volume of the gas and T is the temperature of the gas in Kelvin.
For two situations, we can use that as,
P₁V₁/T₁= P₂V₂/T₂
P₁ = 795 mm Hg
V₁ = 642 mL
T₁ = (273 + 23.5) K = 296.5 K
P₂ = ?
V₂ = 957 mL
T₂ = (273 + 31.7) K = 304.7 K
From substitution,
795 mm Hg x 642 mL / 296.5 K = P₂ x 957 mL / 304.7 K
P₂ = 548.072 mm Hg
760 mmHg = 1 atm
548.072 mm Hg = 1 atm x (548.072 mmHg / 760 mmHg)
= 0.721 atm
Pressure of Oxygen gas is 0.721 atm.
Answer is "A"
Here, we made an assumption that oxygen gas has an ideal gas behavior.
Answer:
Which is higher? - 1979571
Explanation:
Answer:
pH = 10.505
Explanation:
Molar mass of Amphetamine ( C9H13N) = 135 g/mol
Given that the concentration of Amphetamine = 225 mg/L
mass of Amphetamine in one Liter = = 0.225 g
Number of moles of Amphetamine in one liter =
= 0.001667 mol
∴ molarity = 0.0017 M
C₉H₁₃N + H₂O --------> C₉H₁₃NH⁺ + OH⁻
I(M) 0.001667 M 0 0
C(M) -x x x
E(M) 0.001667 - x x x
Pkb = -log Kb = 4.2
∴ Kb = 6.309 x 10⁻⁵
Kb = 6.309 x 10⁻⁵
Equilibrium constant = [C₉H₁₃NH⁺][OH⁻]/ [C₉H₁₃N]
6.309 x 10⁻⁵ = x² / 0.001667-x
where 0.001667 -x ≅ 0.001667
Then;
x² = 6.309 x 10⁻⁵ × 0.001667
x² = 1.0517103 × 10⁻⁷
x =
x = 0.00032 M
x = [OH-] = 0.00032 M
∴ pOH = -log [OH-]
pOH = -log (0.00032)
pOH =3.495
pH = 14 - 3.495
= 10.505