Answer: 0.67 moles of 
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
According to stoichiometry:
3 moles of 
is produced by 2 moles of
Thus 1 mole of is produced by= 
of
Thus 0.67 moles of are required to produce 28.3 g of
Answer:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:
Thus, by combining them, we obtain:
Which is related to the general line equation:
Whereas:
It means that we answer to the blanks as follows:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
Regards!
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
<span>(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)
</span><span>(B)hydrochloric acid + sodium hydroxide
</span><span>HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)
</span><span>(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> </span>AgCl(s) +Ca(NO3)2(aq)
<span>(D)sodium chloride + silver nitrate
</span>NaCl(aq) + AgNO3(aq) ----> AgCl(s) +NaNO32(aq)
AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.
Answer:
Part A
K = (K₂)²
K = (K₃)⁻²
Part B
K = √(Ka/Kb)
Explanation:
Part A
The parent reaction is
2Al(s) + 3Br₂(l) ⇌ 2AlBr₃(s)
The equilibrium constant is given as
K = [AlBr₃]²/[Al]²[Br₂]³
2) Al(s) + (3/2) Br₂(l) ⇌ AlBr₃(s)
K₂ = [AlBr₃]/[Al][Br₂]¹•⁵
It is evident that
K = (K₂)²
3) AlBr₃(s) ⇌ Al(s) + 3/2 Br₂(l)
K₃ = [Al][Br₂]¹•⁵/[AlBr₃]
K = (K₃)⁻²
Part B
Parent reaction
S(s) + O₂(g) ⇌ SO₂(g)
K = [SO₂]/[S][O₂]
a) 2S(s) + 3O₂(g) ⇌ 2SO₃(g)
Ka = [SO₃]²/[S]²[O₂]³
[SO₃]² = Ka × [S]²[O₂]³
b) 2SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
Kb = [SO₃]²/[SO₂]²[O₂]
[SO₃]² = Kb × [SO₂]²[O₂]
[SO₃]² = [SO₃]²
Hence,
Ka × [S]²[O₂]³ = Kb × [SO₂]²[O₂]
(Ka/Kb) = [SO₂]²[O₂]/[S]²[O₂]³
(Ka/Kb) = [SO₂]²/[S]²[O₂]²
(Ka/Kb) = {[SO₂]/[S][O₂]}²
Recall
K = [SO₂]/[S][O₂]
Hence,
(Ka/Kb) = K²
K = √(Ka/Kb)
Hope this Helps!!!
