Answer:
112.2L
Explanation:
Volume (V) = 300g
Temperature (T) = 822K
Pressure (P) = 0.9atm
using the ideal gas equation;
Molar gas constant (R) =
Mole (n) = Molar mass of Mercury = 200.59g/mol
= 1.496mol
Now, the volume can be calculated;
V =
∴Volume of mercury =
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:
Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L
Explanation:
STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.
According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .
Molar mass of Lithium Oxide = 29.8 g/mol
= 2(6.9) + 15.99 = 29.8
Mass of 1 mole = 29.8 g
1 mole occupies, Volume = 22.4 L
29.8 g occupies, V = 22.4 L
1 g occupies ,V =
1 g occupies ,V = 0.7516 L
10 g tex]Li_{2}O[/tex] occupies ,V = L
V = 7.52 L
So, volume occupied by Lithium Oxide At STP is 7.52 L