Atmospheric pressure at sea level is equal to a column of 760 mm Hg. Oxygen makes up 21 percent of the atmosphere by volume. The
1 answer:
<u>Answer:</u> The partial pressure of oxygen is 160 mmHg
<u>Explanation:</u>
We are given:
Percent of oxygen in air = 21 %
Mole fraction of oxygen in air =
To calculate the partial pressure of oxygen, we use the equation given by Raoult's law, which is:
where,
= partial pressure of oxygen = ?
= total pressure of air = 760 mmHg
= mole fraction of oxygen = 0.21
Putting values in above equation, we get:
Hence, the partial pressure of oxygen is 160 mmHg
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Answer:
a) 35.485 moles of Al(SO)
b) 35.485 moles of S atoms
c) Al atoms
d) 567.723 g of O
Explanation:
Let's define the following terms :
1 mol = elemental units
For example :
1 mol of oxygen atoms = oxygen atoms
Now, our compound has the following formula
Al(SO)
Where Al is aluminium
S is sulfur
And O is oxygen
All the subscripts are 1 so we can say the following :
1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O
In terms of moles :
1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O
The molar masses of Al, S and O are
If we sum all the molar masses =
Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.
1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.
Now we can calculate a),b),c) and d)
For a)
75.046 g of Al(SO) = 1 mol of Al(SO)
2663 g of Al(SO) = x
2.663 kg of Al(SO) contains 35.485 moles of Al(SO)
b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms
We have 35.485 moles of Al(SO) molecules so
We have 35.485 moles of S atoms
And 35.485 moles of Al atoms
If 1 mol =
35.485 moles of Al have Al atoms
d) 75.046 g of Al(SO) contains 15.999 g of O
2663 g of Al(SO) contains x g of O
x = 567.723 g of O