Answer: The moles of gas present in the cylinder is 0.34 moles.
Explanation:
Given:
= 2.7 atm,
= 3.1 L,
= 300 K
= ?,
= 9.4 L,
= 610 K
Formula used to calculate the final temperature is as follows.

Substitute the values into above formula as follows.

Now, moles present upon heating the cylinder are as follows.

Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.
Answer) D. ketone. Im not sure if its right tho
Answer:
Graphs should be titled.
Although the bunnies feed foxes, if there are TOO many foxes, they don’t go well; so that looked cyclic to me. It is a good argument.
STUDY VS GRADES is INCREASINGLY
OBVIOUS! I already know what hummingbirds like. Rain can help wash the smoke from forest fires from the air; do you pick one, OK?
Don’t be nervous - study more. I’ve had a great many students, and they not only survived my classes - most THRIVED!
Explanation:
The answer would be A) cells!!
Answer:
5.004kg
Explanation:
Combustion of carbon
C+O2=CO2
from the relationship of molar ratio
mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)
mass of carbon =1000kg
atomic mass of carbon =12
volume of CO2 produced=1000×22.4/12
volume of CO2 produced =1866.6dm3
from the combustion reaction equation provided
CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)
applying the same relationship of molar ratio
no of mole of CO2=no of mole of urea
therefore
vol of CO2\22.4=mass of urea/molar mass of urea
molar mass of urea=60.06g/mol
from the first calculation
vol of CO2=1866.6dm3
mass of urea=1866.6×60.06/22.4
mass of urea=5004.82kg