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Tems11 [23]
2 years ago
13

What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..

Chemistry
1 answer:
Aloiza [94]2 years ago
3 0

Molar mass of NH_3

\\ \sf\longmapsto 14u+3(1u)

\\ \sf\longmapsto 14u+3u

\\ \sf\longmapsto 17g/mol

We know.

No of moles=Given mass/Molar mass

\\ \sf\longmapsto Given\;Mass=17(5)

\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g

Now

Lets write the balanced equation

\\ \sf\longmapsto N_2+3H_2=2NH_3

  • There is 2moles of Ammonia
  • 3moles of H_2
  • 1mole of N_2

Now

\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}

For Hydrogen

\\ \sf\longmapsto 3\times 6.023\times 10^{23}

\\ \sf\longmapsto 18.069\times 10^{23}

\\ \sf\longmapsto 1.8\times 10^{22}molecules

For Ammonia

\\ \sf\longmapsto 2\times 6.023\times 10^{23}

\\ \sf\longmapsto 12.046\times 10^{23}

\\ \sf\longmapsto 1.2\times 10^{22}molecules

For Nitrogen

\\ \sf\longmapsto 1\times 6.023\times 10^{23}

\\ \sf\longmapsto 6.023\times 10^{23}molecules

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3 years ago
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
user100 [1]

Answer:

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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Explanation:

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