Answer:
92.87 g.
Explanation:
∵ The percentage yield = (actual yield/theoretical yield)*100.
- We need to calculate the theoretical yield:
From the balanced reaction:
<em>PCl₃ + Cl₂ → PCl₅,</em>
It is clear that 1 mol of PCl₃ reacts with 1 mol of Cl₂ to produce 1 mol of PCl₅.
- We need to calculate the no. of moles of 73.7 g PCl₃:
n = mass/molar mass = (73.7 g)/(137.33 g/mol) = 0.536 mol.
<u><em>Using cross multiplication:</em></u>
1 mol of PCl₃ produce → 1 mol of PCl₅, from stichiometry.
∴ 0.536 mol of PCl₃ produce → 0.536 mol of PCl₅.
∴ The mass of PCl₅ (theoretical yield) = (no. of moles) * (molar mass) = (0.536 mol)*(208.24 g/mol) = 111.62 g.
<em>∵ The percentage yield = (actual yield/theoretical yield)*100.</em>
The percentage yield = 83.2%, theoretical yield = 111.62 g.
∴ The actual yield of PCl₅ = (The percentage yield)(theoretical yield)/100 = (83.2%)(111.62 g)/100 = 92.87 g.
Answer:
-767,2kJ
Explanation:
It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:
1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ
2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ
4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ
5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ
The sum of (4) - (2) produce:
6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ
(6) + 4×(3):
7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ
(7) - 2×(1):
8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ
(8) - 2×(5):
9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= <em>-767,2kJ</em>
I hope it helps!
<span> As we know that MgI2 (magnesium iodide) when dissociated it gives more ions than the KI so it has more boiling point as its boiling point is high it means that it boils more so it has low vapor pressure and freezing point
On the other hand as we know that KI dissociates into two ions so so it has high freezing and vapor pressure
hope it helps</span>
<span>
It makes sense that an inner shell electron would be tougher to remove
than a valence electron because the inner shell electron is closer to
the positive nucleus of the atom. Seeing as an electron caries a
negative charge it would be too attracted to the positive core to leave
readily. Also, the inner shell electrons are constantly repelling
electrons outside of it's energy level (however the reason these
electrons outside innershell energy levels don't simply fly away is the
charge of the positive core overcomes the smaller charges of the
comparably negligible inner shell electrons, but that repulsion is still
there so keep that in mind) </span>
