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2 years ago

What is the approximate partial pressure of oxygen at 1500 m?

Chemistry

1 answer:

elena55 [62]2 years ago

3 0

The approximate partial pressure of oxygen at 1500 m is 0.18 atm.

<h3>Pressure of the air</h3>

From barometric pressure table;

1500 m = 85 kPa

1 kPa = 0.0099 atm

85 kPa = ?

= 0.84 atm

Oxygen percentage in air = 21%

<h3>Partial pressure of oxygen in air</h3>

P = 0.21 x P(total)

P = 0.21 x 0.84 atm

P = 0.18 atm

Thus, the approximate partial pressure of oxygen at 1500 m is 0.18 atm.

Learn more about partial pressure here: brainly.com/question/19813237

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3 years ago

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3 years ago

A sample of propane(c3h8)has 3.84x10^24 H atoms.

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Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

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________________________________________

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= 3 × 12 + 8 × 1

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= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

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$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

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2 years ago

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3 years ago

Read 2 more answers

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$b.$ $obtain$