The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
<h3>Final volume (V₂) =? </h3>
- The final volume of the gas can be obtained by using the combined gas equation as illustrated below:
Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023
<h3>V₂ = 15 L </h3>
Therefore, the final volume of the gas is 15 L
Learn more: brainly.com/question/25547148
5.6 Al(OH)3
5.6 Al, 16.8 O, 16.8 H
16.8 mols of oxegyn in 5.6 mols of Al(OH)3
Answer: 4.41 atm
Explanation:
Given that,
Original pressure of oxygen gas (P1) = 5.00 atm
Original temperature of oxygen gas (T1) = 25°C
[Convert 25°C to Kelvin by adding 273
25°C + 273 = 298K
New pressure of oxygen gas (P2) = ?
New temperature of oxygen gas (T2) = -10°C
[Convert -10°C to Kelvin by adding 273
-10°C + 273 = 263K
Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law
P1/T1 = P2/T2
5.00 atm /298K = P2/263K
To get the value of P2, cross multiply
5.00 atm x 263K = 298K x V2
1315 atm•K = 298K•V2
V2 = 1315 atm•K / 298K
V2 = 4.41 atm
Thus, the new pressure inside the canister is 4.41 atmosphere
S4O5
S3O
SeF6
N4S5
CCl9
All numbers should be subscripts
