<u>Answer:</u> The volume of the container needed is 554.6 L
<u>Explanation:</u>
To calculate the volume of the gas, we use the equation given by ideal gas which follows:

where,
P = pressure of the gas = 3.0 atm
V = Volume of the gas = ? L
T = Temperature of the gas = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles of propane gas = 68.0 moles
Putting values in above equation, we get:

Hence, the volume of the container needed is 554.6 L
If it is 60 Celsius that would conver to fare height by means of this equation; (1.8*60)+32°F
Which would come out to.... 140° Fahrenheit... Hardly seems like chilly conditions.
Ncomplete combustion of<span> fossil </span>fuels<span>; forest fires// heavy traffic ... NS: </span>oxidation<span> of H2S </span>gas<span>from </span>decay<span> of </span>organic matter<span> & volcanic activity ... primary pollutant; </span>burning<span> of </span>sulfur containing<span>fossil </span>fuels<span>, </span>coal<span> containing ... HS: </span>combustion of<span> fossil </span>fuel<span>, industrial plants that </span>produce<span> smoke, ash, dust ..... </span>photochemical<span> smog.</span>
I need the graph to answer your question
Answer:
The theoretical yield of urea = <u>120.35kg</u>
The percent yield for the reaction = <u>72.70%</u>
Explanation:
Lets calculate -
The given reaction is -
→
Molar mass of urea
= 60g/mole
Moles of
=
(since
)
= 4011.76 moles
Moles of
= 
= 
= 2386.36 moles
Theoritically , moles of
required = double the moles of 
but ,
, the limiting reagent is 
Theoritical moles of urea obtained = 
= 
Mass of 2005.88 mole of
=
= 120352.8g

= 120.35kg
Therefore , theroritical yeild of urea = 120.35kg
Now , Percent yeild = 
72.70%
Thus , the percent yeild for the reaction is 72.70%