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2 years ago

0.7 L of H2 enters a reaction at standard temperature and pressure.

Chemistry

1 answer:

ASHA 777 [7]2 years ago

6 0

Answer:

1.88 × 10²³ particles

Explanation:

Given data:

Volume of H₂ = 0.7 L

Number of particles at STP = ?

Solution:

First of all we will calculate the number of moles of H₂.

PV = nRT

n = PV / RT

n = 1 atm . 7 L / 0.0821 L. atm. mol⁻¹. k⁻¹ × 273.15 K

n = 7 atm. L / 22.43 L. atm. mol⁻¹

n = 0.312 mol

it is known that,

2 g H₂ = 1 mol = 6.022 × 10²³ particles

For 0.312 mol

0.312 × 6.022 × 10²³ particles

1.88 × 10²³ particles

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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
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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.

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2 years ago

Why is cellulose not a source of nutrients for humans? Humans cannot absorb the hydrolyzed form of cellulose. Humans, and most v

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2 years ago

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1 year ago

A small hole in the wing of a space shuttle requires a 20.7-cm² patch.

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The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

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or, 1 cm² = (0.00001 km)²

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20.7 cm² = 20.7 × 10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

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