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statuscvo [17]
1 year ago
5

Calculate the energy required to produce 6.0 moles of Cl2O7 using the following reaction 2 Cl2 + 7 O2 + 130kcal -> 2 Cl207

Chemistry
1 answer:
PilotLPTM [1.2K]1 year ago
5 0

Answer:

455 Kcal

Explanation:

2Cl2(g) + 7O2(g) + 130kcal → 2Cl2O7(g)

Rearranging we get,

2Cl2(g) + 7O2(g)  → 2Cl2O7(g) Δ H = 130 kcal . mol⁻¹

So for per mol reaction will be as above.

In case of 7 mols of product, we need 7/2 mole ratio x 130 = 455 Kcal

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Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s
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Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.
6 0
3 years ago
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