First, we need to compute the mass of oxygen found in 100 grams of saltpeter:
mass of oxygen = 100 - (mass of potassium + mass of nitrogen)
= 100 - (38.67 + 13.86)
= 100 - 52.53
mass of oxygen in 100 grams saltpeter = 47.47 grams
Now, we can use cross multiplication to find the mass of oxygen in 328 grams saltpeter as follows:
mass of oxygen = (328 x 47.47) / 100 = 155.7016 grams
Answer:
= 0.28M
Explanation:
data:
volume = 0.250 L
= 0.250dm^3 ( 1litre = 1dm^3)
moles = 0.70 moles
Solution:
molarity = 
= 0.70 / 0.250
molarity = 0.28 M
Answer:
(a) The proportion of dry air bypassing the unit is 14.3%.
(b) The mass of water removed is 1.2 kg per 100 kg of dry air.
Explanation:
We can express the proportion of air that goes trough the air conditioning unit as
and the proportion of air that is by-passed as
, being
.
The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

Replacing the first equation in the second one we have

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.
It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.
The water removed of every 100 kg of dry air is

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).
100 * 0.012 = 1.2 kgW