Answer:
11.94 grams of carbon dioxide were originally present.
19.94 grams of krypton can you recover.
Explanation:
Mass of carbon dioxide gas = x
Mass of krypton gas = y
x + y = 31.7 g
Moles of carbon dioxide gas = 
Moles of krypton gas = 
Mole fraction of krpton =
Total pressure of the mixture = P = 0.665 atm
Partial pressure of carbon dioxide gas = p
Partial pressure of krypton gas before removal of carbon dioxide gas = p'
Partial pressure of krypton gas after removal of carbon dioxide gas = p'' = 0.309 atm
p' = p'' = 0.309 atm
0.665 atm = p + 0.309 atm
p = 0.665 atm - 0.306 atm = 0.359 atm
Partial pressure of krypton can also be given by :



..[2]
Solving [1] and [2]:
x = 11.94 g
y = 19.76 g
11.94 grams of carbon dioxide were originally present.
19.94 grams of krypton can you recover.
I think the answer is that is lights up?
Answer:
A
Explanation:
Because in 1 second, line A was faster than line B.
Answer:
58g
Explanation:
In order to solve this problem, you must take a look at the solubility graph for potassium nitrate.
Now, the solubility graph shows you how much solute can be dissolved per 100g of water in order to make an unsaturated, a saturated, or a supersaturated solution.
You're looking to make a saturated potassium nitrate solution using
50g of water at 60∘C. Your starting point will be to determine how much potassium nitrate can be dissolved in 100g of water at that temperature in order to have a saturated solution.
As you can see, the curve itself represents saturation.
If you draw a vertical line that corresponds to 60∘C and extend it until it intersects the curve, then draw a horizontal line that connects to the vertical axis, you will find that potassium has a solubility of about
115g per 100g of water. Your answer is 58g of potassium nitrate