Answer:
It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.
Explanation:
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
So, in this case:
- P= 1 atm
- V= 855 L
- n= ?
- R= 0.082

- T= 273.15 K
Replacing:
1 atm* 855 L= n* 0.082
* 273.15 K
Solving:

n= 38.17 moles
Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?

mass= 1,068.76 grams
<u><em>
It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.</em></u>
Remembering that
d = m ÷ v
d = ?
m = 89 g
v = 10 cm³
Therefore:
d = 89 ÷ 10
d = 8,9 g÷cm³
Answer:
1455.6
Explanation: you first convert 2250ml to l by dividing by 1000 so you get 2.25l then you use Boyles law which is p1v1=p2v2 then insert values
35.75*100=p2*2.25 then divide both sides by 2.25 then you get 1455.6