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Vitek1552 [10]
2 years ago
8

A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

ined 3.223 g. It is again evacuated and then filled with a gas of unknown molar mass. When reweighed, the flask is found to have gained 8.105 g.
Chemistry
1 answer:
Elden [556K]2 years ago
8 0

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

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vitfil [10]
<span>Percentage by mass is the amount in mass of a component in a mixture per 100 unit of mass of the total mixture. Percentage by mass is the same as  %w/w. We can determine this by dividing the mass of the solute with the total mass of the mixture. However, from the problem statement, we are given the volume of the water so there is a need to convert this value to mass by using the density of water. We calculate as follows:

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2 years ago
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blondinia [14]

2, 4, 1

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6 0
2 years ago
What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.
german
When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

PV = nRT

when P is the pressure =  1.1 atm

and V is the volume = 14.5 L 

n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

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∴ n = 1.1 * 14.5 / (0.0821*285.5)

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so we can easily get the mass of CaSO3:

when mass = moles * molar mass

and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol


∴ mass = 0.68 moles* 120 g/mol = 81.6 g
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Like many other organic solvents we will use this semester, ethanol is flammable and caution needs to be exercised when heating
laiz [17]

Answer:

See explanation

Explanation:

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