Mass % of nitrogen = mass of nitrogen*100 / total mass
= 14*100 / (1+ 14 + 32)
= 14*100 / 47
= 29.7 %
You have to find the gram formula mass of C6H6 then do mass (g) = mol x GFM
Answer:
2666.7 hours
Explanation:
The key to solve this problem is that we are given the propane gas consumed in one hour by giving us the information of the volume consumed at 1 atm, 298 K (25 +273). Using the gas law we can calculate the rate of consumption of propane per hour, and from here we can calculate its mass and converting it to gallons and finally diving the 400 gallos by this number.
PV = nRT ∴ n = PV/RT
n = 1 atm x 165 L/ (0.08206 Latm/kmol x 298 K ) = 6.75 mol propane
Mass propane :
6.75 mol x 44 g/mol = 296.88 g
convert this to Kg:
296.88 g/ 1000 g/Kg = 0.30 Kg
calculate the volume in liters this represents by dividing by the density:
0.30 Kg / 0.5077 Kg/L = 0.59 L
changing this to gallons
0.59 L x 1 gallon/3.785 L = 0.15 gallon
and finally calculate how many hours the 400 gallons propane tank will deliver
400 gallon/ 0.15 gallon/hr = 2666.7 hr
Answer:
The volume occupied by 2.34 grams of CO2 gas at STP is 1.18 L
Answer:
solubility of X in water at 17.0 
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = 
= 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
