Answer:
8 m
Explanation:
3.0 x 10*8 divided by 3.75 x 10*7 = 8 m
In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.
O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg) = 0.13
CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044
Answers: O2 = 0.13
CO2 = 0.044
Answer:
The 12L helium tank pressurized to 160 atm will fill <em>636 </em>3-liter balloons
Explanation:
It is possible to answer this question using Boyle's law:
Where P₁ is the pressure of the tank (160atm), V₁ is the volume of the tank (12L), P₂ is the pressure of the balloons (1atm, atmospheric pressure) And V₂ is the volume this gas will occupy at 1 atm, thus:
160atm×12L = 1atm×V₂
V₂ = 1920L
As the tank will never be empty, the volume of the gas able to fill balloons is the total volume minus 12L, thus the volume of helium able to fill balloons is:
1920L - 12L = 1908L
1908L will fill:
1908L×
= <em>636 balloons</em>
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I hope it helps!
Answer:
If there is 0.66 moles of iron(III)oxide produced, there reacte 0.99 moles of oxygen (O2)
Explanation:
Step 1: Data given
Number of moles iron (III) oxide (Fe2O3) = 0.66 moles
Step 2: The balanced equation
4Fe + 3O2 → 2Fe2O3
Step 3: Calculate moles of oxygen (O2)
For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3
For 0.66 moles Fe2O3 produced, we need 3/2 * 0.66 = 0.99 moles of O2
If there is 0.66 moles of iron(III)oxide produced, there reacte 0.99 moles of oxygen (O2)
