Answer:
0.414 mole (3 sig. figs.)
Explanation:
Given grams, moles = mass/formula weight
moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)
≅ 0.414 mole (3 sig. figs.)
The partial pressure of oxygen given the total barometric pressure is : 108.15 mmHg
<u>Given data : </u>
Total barometric pressure = 515 mmHg
Assuming oxygen percentage = 21%
Barometric pressure dry at 37°C
<h3 /><h3>Determine the partial pressure of oxygen </h3>
Applying the relation below
Partial pressure = oxygen percentage * Barometric pressure
= 21% * 515 mmHg
= 108.15 mmHg
Hence we can conclude that the partial pressure of oxygen is 108.15 mmHg.
Learn more about Partial pressure : brainly.com/question/1835226
Nitrogen and oxygen are the most prevalent in the atmosphere.
The ratio of effusion rates for the lightest gas H₂ to the heaviest known gas UF₆ is 13.21 to 1
<h3>What is effusion?</h3>
Effusion is a process by which a gas escapes from its container through a tiny hole into evacuated space.
Rate of effusion ∝ 1/√Ц, (where Ц is molar mass)
Rate H₂ = 1/√ЦH₂
Rate UF₆ = 1/√ЦUF₆
Therefore, Rate H₂/ Rate UF₆ = √ЦH₂/√ЦUF₆
ЦH₂= 2.016 g/mol
ЦUF₆= 352.04 g/mol
Rate H₂ / Rate UF₆ = √352.04/√2.016 = 18.76/1.42
Rate H₂ / Rate UF₆ = 13.21
Therefore, H₂ is lower mass than UF₆. Thus H₂ gas will effuse 13 times more faster than UF₆ because the most probable speed of H₂ molecule is higher; therefore, more molecules escapes per unit time.
learn more about effusion rate: brainly.com/question/28371955
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