Answer:
-1268 kJ
Explanation:
The molar enthalpy of combustion of C6H6 is -3170 kJ/mol
2C6H6(g) + 15O2(g) → 12CO2(g) + 6H2O(g)
Therefore;
when 1 mole of C6H6 undergoes combustion, an energy of -3170 kJ/mol is released.
Therefore; for 0.40 mole
= 0.4 × -3170 kJ
= -1268 kJ
Answer:
Explanation:
Given parameters:
Concentration of H₃O⁺ = 5.6 x 10⁻²M
Solution:
To solve for the concentration of H₃O⁺ in the solution, we simply use the expression below:
pH = -log₁₀[H₃O⁺]
where [H₃O⁺] = 5.6 x 10⁻²M is the concentration of H₃O⁺
pH = -log₁₀[5.6 x 10⁻²] = - x -1.25 = 1.25
Answer:
1.21 g of Tris
Explanation:
Our solution if made of a solute named Tris
Molecular weight of Tris is 121 g/mol
[Tris] = 100 mM
This is the concentration of solution:
(100 mmoles of Tris in 1 mL of solution) . 1000
Notice that mM = M . 1000 We convert from mM to M
100 mM . 1 M / 1000 mM = 0.1 M
M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris
0.1 M = mmoles of Tris / 100 mL
mmoles of Tris = 100 mL . 0.1 M → 10 mmoles
We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol
And now we determine the mass of solute, by molecular weight
0.010 mol . 121 g /mol = 1.21 g
Answer:
ΔG° = -533.64 kJ
Explanation:
Let's consider the following reaction.
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
The standard Gibbs free energy (ΔG°) can be calculated using the following expression:
ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)
where,
ni are the moles of reactants and products
ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products
ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)
ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)
ΔG° = -533.64 kJ
Answer:
Water can dissolve salt because the positive part of water molecules attracts the negative chloride ions and the negative part of water molecules attracts the positive sodium ions. The amount of a substance that can dissolve in a liquid (at a particular temperature) is called the solubility of the substance.
Explanation:
