Answer:
1219.5 kj/mol
Explanation:
To reach this result, you must use the formula:
ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).
The BE values are:
BE C = C: 839 kj / mol
BE C-H: 413 Kj / mol
BE O = O: 495 kj / mol
BE C = O = 799 Kj / mol
BE O-H = 463 kj / mol
Now you must replace the values in the above equation, the result of which will be:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol
Answer is: a lower freezing point has solution of K₂SO₄.
Change in freezing
point from pure solvent to solution: ΔT =i · Kf · b.<span>
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per
kilogram of solvent.
i - </span>Van't
Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
Answer:
748 torr
Explanation:
mmHg and torr are equivalent so, you'll have 748 torr.
Answer:
4.34.
Explanation:
<em>∵ pH = pKa + log [salt]/[Acid]</em>
∴ pH = - log(Ka) + log [salt]/[Acid]
∴ pH = - log(6.8 x 10⁻⁵) + log(0.75)/(0.50)
<em>∴ pH = 4.167 + 0.176 = 4.343 ≅ 4.34.</em>
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