Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
Answer:
it would be option A
Explanation:
This is becuase if you look at the chart you can see tyhat the group of rats that got feed to vitamans did gain more wati then the ones on the normal diet.
In order to decrease the chances of deception in acquiring a handwriting sample, what the law enforcement do is that, they ask they person to write as little as possible, write the passage only once and also gather the material that consist some of the similar words and phrases as the suspected document. Answer for this would be all of the above.
Answer:
half lives passed=5
given sample=90g
sample left=2.8125g
Explanation:
no. of half lives=total time/half life
no.=19days/3.8days
no.=5 days
after 5 half lives sample left=2.8125g