A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increa
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Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+
⇄
Formation constant Kf
Kf = / ( [
][
] ) = 5.0 × 10¹⁰
Now,
[] =
; ∝₄ = 0.35
so the equilibrium is;
+
⇄
+ 4H⁺
Given that; =
{ 1 mol
reacts with 1 mol
}
so at equilibrium, =
= x
∴
+
⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [
][
] ) =
/ ( [
][
] ) = 5.0 × 10¹⁰
⇒ / ( [
][
] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
<u>Answer:</u> The percentage yield of HF is 73.36 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
....(1)
For calcium fluoride:
Given mass of calcium fluoride = 6.25 kg = 6250 g (Conversion factor: 1 kg = 1000 g)
Molar mass of calcium fluoride = 78.07 g/mol
Putting values in above equation, we get:
For the given chemical reaction:
By Stoichiometry of the reaction:
1 mole of calcium fluoride produces 2 moles of hydrofluoric acid
So, 80.05 moles of calcium fluoride will produce = of hydrofluoric acid
Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:
Moles of of hydrofluoric acid = 160.1 moles
Molar mass of hydrofluoric acid = 20.01 g/mol
Putting values in equation 1, we get:
To calculate the percentage yield of hydrofluoric acid, we use the equation:
Experimental yield of hydrofluoric acid = 2.35 kg
Theoretical yield of hydrofluoric acid = 3.20 kg
Putting values in above equation, we get:
Hence, the percentage yield of HF is 73.36 %