Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
According to the reversible reaction equation:
2Hi(g) ↔ H2(g) + i2(g)
and when Keq is the concentration of the products / the concentration of the reactants.
Keq = [H2][i2]/[Hi]^2
when we have Keq = 1.67 x 10^-2
[H2] = 2.44 x 10^-3
[i2] = 7.18 x 10^-5
so, by substitution:
1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2
∴[Hi] = 0.0033 M
1)They are all take up space.
2)They all have mass.
3)They are all solids.
Because of 1 and 2, they are all matter.
Answer is: the solution is saturated.
Solubility of potassium chloride (KCl) on 20°C is 34.2 grams in 100 grams of water, so in 200 grams of water will dissolve two times more salt (68.4 g).
Saturated solution contains the maximum concentration of a solute dissolved in the solvent (usually water) and if extra solute is add to saturated solution, that solute will not dissolve.
The amount of solute that can be dissolved in a solvent depends of chemical composition, temperature and pressure.
