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3 years ago

If .758 moles of gas occupy a volume of 80.6L, how many moles will occupy a volume of 270.9L?

Chemistry

1 answer:

egoroff_w [7]3 years ago

4 0

Answer:

n₂ = 2.55 mol

Explanation:

Given data:

Initial number of moles = 0.758 mol

Initial volume = 80.6 L

Final volume = 270.9 L

Final number of moles = ?

Solution:

Formula:

V₁/n₁ = V₂/n₂

V₁ = Initial volume

n₁ = initial number of moles

V₂ = Final volume

n₂ = Final number of moles

now we will put the values in formula.

V₁/n₁ = V₂/n₂

80.6 L / 0.758 mol = 270.9 L/ n₂

n₂ = 270.9 L× 0.758 mol / 80.6 L

n₂ = 205.34 L.mol /80.6 L

n₂ = 2.55 mol

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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

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

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

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

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

= 14.25 × 10²³ 

________________________________________

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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

=> 4. 75 × 10²³ molecules of Propane weigh =

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

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

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

= 34.72 g

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