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o-na [289]
3 years ago
5

PLEASE HELP!!!

Chemistry
1 answer:
NemiM [27]3 years ago
7 0

Answer:deez nuggars

Explanation:

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If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give
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The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

  • The maximum height attained by nitrogen gas molecule = h
  • The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

  • The potential energy at height h = P.E = 6.381\times 10^{-21} J
  • Molar mass of nitrogen gas =  28.0134 g/mol
  • Mass of nitrogen gas molecule = m

m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg

  • The acceleration due to gravity = g = 9.8 m/s^2
  • The maximum height attained by nitrogen gas molecule = h
  • The potential energy is given by:

P.E=mgh

6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Learn more about the average kinetic energy of gas particles here:

brainly.com/question/16615446?referrer=searchResults

brainly.com/question/6329137?referrer=searchResults

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Answer:

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trial 3: 5.19

I think but the equation for sloving percent error is:

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