Answer: There are now 2.07 moles of gas in the flask.
Explanation:
P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = ?
n = number of moles = 1.9
T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.
P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = 49.8 L
n = number of moles = ?
T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
Thus the now the container contains 2.07 moles.
3 elements are present in HCOOH - hydrogen, oxygen, and carbon.
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>