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olya-2409 [2.1K]
3 years ago
15

What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:

Chemistry
1 answer:
Margarita [4]3 years ago
6 0

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
  • n = 2.5 mol

Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

  • 1.67 mol * 65 g/mol = 108.6 g
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Answer:

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Explanation:

Hello,

In this case, for the calculation of the standard entropy of liquid lead at 500 °C (773.15 K), starting by solid lead 298.15 K we need to consider three processes:

1. Heating of solid lead at 298.15 K to 600.55 K (melting point).

2. Melting of solid lead to liquid lead.

3. Heating of liquid lead at 600.55 K (melting point) to 773.15 K.

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s_{Pb(l),500\°C}=s_{Pb(s),298.15K}+s_1+s_2+s_3

Whereas each entropy is computed as follows:

s_1=\int\limits^{600.55K}_{298.15K} {\frac{22.13 + 0.01172 T  + 1.00 x 10^{-5} T^2}{T} } \, dT =20.4\frac{J}{mol*K}\\\\\\s_2=\frac{4770\frac{J}{mol} }{600.55K}= 7.94\frac{J}{mol*K}\\\\\\s_3=\int\limits^{773.15K}_{600.55K} {\frac{32.51-0.00301T}{T} } \, dT=7.69\frac{J}{mol*K}

Therefore, the standard entropy of liquid lead at 500 °C turns out:

s_{Pb(l),500\°C}=64.80+20.4+7.94+7.69\\\\s_{Pb(l),500\°C}=100.83\frac{J}{mol*K}

Best regards.

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