The answer is always found if you think
so i would pick b
The solubility of carbon dioxide in the soda container when the soda can is opened is decreased. This is because the pressure is decreased resulting to the decreased in solubility. The law that governs this is the Henry's law.
Answer:
ΔHorxn = - 11.79 KJ
Explanation:
2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )
The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol
From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,
1 mol = 22.4l
x mol = 2.67l
Upon cross multiplication and solving for x;
x = 2.67 / 22.4 = 0.1192 mol
0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.
Amount of heat is given as;
ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)
Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.
ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)
ΔHorxn = - 47.17kj + 35.38kj
ΔHorxn = - 11.79 KJ
Answer:
91.26 g
Explanation:
Given data:
Mass of PF₃ = 180 g
Mass of F₂ required = ?
Solution:
Chemical equation:
P₄ + 6F₂ → 4PF₃
Moles of PF₃:
Number of moles = mass/ molar mass
Number of moles = 180 g/ 88 g/mol
Number of moles = 2.05 mol
Now we will compare the moles of PF₃ with F₂.
PF₃ : F₂
4 : 6
2.05 : 6/4×2.05 = 3.075
Mass of F₂:
Mass of F₂ = moles × molar mass
Mass of F₂ = 3.075 mol × 38 g/mol
Mass of F₂ = 116.85 g
If reaction yield is 78.1%:
116.85 /100 ×78.1 = 91.26 g
