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Mars2501 [29]
3 years ago
6

What is the oxidation number of Nitrogen in HNO2? +1 -1 +3 -3

Chemistry
1 answer:
WITCHER [35]3 years ago
5 0
I believe the correct answer from the choices listed above is the third option. The <span>oxidation number of Nitrogen in HNO2 would be +3. It is calculated as follows:

1 + x + (-2)(2) = 0
x = +3

Hope this answers the question. Have a nice day.</span>
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The heat of vaporization for benzaldehyde is 48.8 kj/mol, and its normal boiling point is 451.0 k. use this information to deter
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The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

Explanation:

  • To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
  • The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
  • The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
  • ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
  • Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
  • ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
  • ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
  • (760 torr /P₂) = 0.01075
  • Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.

So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

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