Mass in kilograms of liquid air required = 0.78 kg
<u>Given that </u>
1 Litre of liquid air contains 1.3 grams of oxygen ( air )
<u />
<u> Determine the a</u><u>mount of liquid air</u><u> in Kg</u>
volume of air given = 600 L
mass of liquid air required = x
1 litre = 1.3 grams
600 L = x
∴ x ( mass of liquid air ) = 1.3 * 600
= 780 g = 0.78 kg
Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg
Learn more about liquid air : brainly.com/question/636295
The given question is incorrect. The correct question is as follows.
If 20.0 g of
and 4.4 g of
are placed in a 5.00 L container at
, what is the pressure of this mixture of gases?
Explanation:
As we know that number of moles equal to the mass of substance divided by its molar mass.
Mathematically, No. of moles = 
Hence, we will calculate the moles of oxygen as follows.
No. of moles = 
Moles of
=
= 0.625 moles
Now, moles of 
= 0.1 moles
Therefore, total number of moles present are as follows.
Total moles = moles of
+ moles of 
= 0.625 + 0.1
= 0.725 moles
And, total temperature will be:
T = (21 + 273) K = 294 K
According to ideal gas equation,
PV = nRT
Now, putting the given values into the above formula as follows.
P =
= 
=
atm
= 3.498 atm
or, = 3.50 atm (approx)
Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.
Your answer is B. 6.02 x 1023 grams
Answer:
a) is correct.......,.....