The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol
<h3>Further explanation</h3>Given
Ratio of the concentrations of the products to the concentrations of the reactants is 22.3
Temperature = 37 C = 310 K
ΔG°=-16.7 kJ/mol
Required
the free energy change
Solution
Ratio of the concentration : equilbrium constant = K = 22.3
We can use Gibbs free energy :
ΔG = ΔG°+ RT ln K
R=8.314 .10⁻³ kJ/mol K
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
The complete balanced equation for the given reaction is expressed as;
Given the following parameters
Mass of CH4 = 5.90×10^−3 g = 0.0059grams
Determine the moles of methane
According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:
Determine the mass of water produced
Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams
This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.
Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:
Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:
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