Answer:
0.0917 mol Co(CrO₄)₃
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
37.3 g Co(CrO₄)₃
<u>Step 2: Identify Conversions</u>
Molar Mass of Co - 58.93 g/mol
Molar Mass of Cr - 52.00 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol
<u>Step 3: Convert</u>
<u />
= 0.091662 mol Co(CrO₄)₃
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃
Last option that is none of above is right answer.
Answer : The partial pressure of the
in the tank in psia is, 32.6 psia.
Explanation :
As we are given 75 %
and 25 %
in terms of volume.
First we have to calculate the moles of
and
.
![\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCO_2%3D%5Cfrac%7B%5Ctext%7BVolume%20of%20%7DCO_2%7D%7B%5Ctext%7BVolume%20at%20STP%7D%7D%3D%5Cfrac%7B75%7D%7B22.4%7D%3D3.35mole)
![\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DN_2%3D%5Cfrac%7B%5Ctext%7BVolume%20of%20%7DN_2%7D%7B%5Ctext%7BVolume%20at%20STP%7D%7D%3D%5Cfrac%7B25%7D%7B22.4%7D%3D1.12mole)
Now we have to calculate the mole fraction of
.
![\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DCO_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DCO_2%7D%7B%5Ctext%7BMoles%20of%20%7DCO_2%2B%5Ctext%7BMoles%20of%20%7DN_2%7D)
![\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DCO_2%3D%5Cfrac%7B3.35%7D%7B3.35%2B1.12%7D%3D0.75)
Now we have to calculate the partial pressure of the
gas.
![\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}](https://tex.z-dn.net/?f=%5Ctext%7BPartial%20pressure%20of%20%7DCO_2%3D%5Ctext%7BMole%20fraction%20of%20%7DCO_2%5Ctimes%20%5Ctext%7BTotal%20pressure%20of%20gas%7D)
![\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}](https://tex.z-dn.net/?f=%5Ctext%7BPartial%20pressure%20of%20%7DCO_2%3D0.75mole%5Ctimes%20300Kpa%3D225Kpa%3D225Kpa%5Ctimes%20%5Cfrac%7B0.145%5Ctext%7B%20psia%7D%7D%7B1Kpa%7D%3D32.625%5Ctext%7B%20psia%7D)
conversion used : (1 Kpa = 0.145 psia)
Therefore, the partial pressure of the
in the tank in psia is, 32.6 psia.
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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