Answer:
d. Sum of product enthalpies minus the sum of reactant enthalpies
Explanation:
The standard enthalpy change of a reaction (ΔH°rxn) can be calculated using the following expression:
ΔH°rxn = ∑n(products) × ΔH°f(products) - ∑n(reactants) × ΔH°f(reactants)
where,
ni are the moles of products and reactants
ΔH°f(i) are the standard enthalpies of formation of products and reactants
Answer:
Zn+H2O ------ ZnO + H2 is the correct answer
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
True, all reversible reactions will reach chemical equilibrium.
Answer:
-122 J/K
Explanation:
Let's consider the following balanced reaction.
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)
We can calculate the standard reaction entropy (ΔS°) using the following expression.
ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r
where,
- η: stoichiometric coefficients of products and reactants
- Sf°r: entropies of formation of products and reactants
ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol
ΔS° = -121.77 J/K ≈ -122 J/K
