Answer:
82.416 g of KNO
₃ is needed to produce 510.0 mL of a 1.6 M KNO
₃ solution.
Explanation:
Since molarity is the number of moles of solute that are dissolved in a given volume, calculated by dividing the moles of solute by the volume of the solution, the following rule of three can be applied: if in 1 L (1,000 mL) of KNO₃ there are 1.6 moles of the compound present, in 510 mL how many moles will there be?

moles= 0.816
Being the molar mass of the elements:
- K: 39 g/mole
- N: 14 g/mole
- O: 16 g/mole
So the molar mass of the compound KNO₃ is:
KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole
Now I can apply the following rule of three: if in 1 mole of KNO₃ there are 101 g, in 0.816 moles how much mass is there?

mass= 82.416 grams
<u><em>82.416 g of KNO
₃ is needed to produce 510.0 mL of a 1.6 M KNO
₃ solution.</em></u>
A decrease in the overall volume of gases namely hydrogen would prevent nuclear fusion in a nebula.
Molarity is moles divided by liters so do .732 divided by .975 liters.
Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g