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:
Spreading out a wave over a larger area just causes the wave strength to weaken, but does not cause gaps to form. Therefore, if you look at photons as waves, spatial gaps never form in light as it travels through free space, no matter how dim it gets.Spreading out a wave over a larger area just causes the wave strength to weaken, but does not cause gaps to form. Therefore, if you look at photons as waves, spatial gaps never form in light as it travels through free space, no matter how dim it gets
Explanation:
Answer is: <span>1.0 mol X left over.
</span>Chemical reaction: X + 2Y → XY₂.
n(X) = 3,0 mol, excess reactant.
n(Y) = 4,0 mol, limiting reagent.
n - amount of substance.
from reaction: n(X) : n(Y) = 1 : 2.
n(X) : 4 mol = 1 : 2.
n(X) = 2 mol, that reacts.
excess of X: 3 mol - 2 mol = 1 mol.
Answer:
This is the electronic configuration of selenium atom.
Explanation:
Selenium is present in group 16.
It is the member of oxygen family.
Its atomic number is 34.
Electronic configuration:
Se₃₄ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴
The noble gas notation is used for the shortest electronic configuration of other periodic table elements.
For example:
The atomic number of Argon is 18, and its electronic configuration is,
Ar₁₈ = 1s² 2s² 2p⁶ 3s² 3p⁶
Abbreviated electronic configuration:
Se₃₄ = [Ar] 4s² 3d¹⁰ 4p⁴
