Considering ideal gas:
PV= RTn
T= 25.2°C = 298.2 K
P1= 637 torr = 0.8382 atm
V1= 536 mL = 0.536 L
:. R=0.082 atm.L/K.mol
:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))
:. n= O.0184 mol
Then,
P2= 712 torr = 0.936842 atm
V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)
:.V2 = 0.4796 L
OR
V2 = 479.6 ml
Explanation:
It is known that one mole of chromium or molar mass of chromium is 51.99 g/mol.
It is given that number of moles is 11.9 moles.
Therefore, calculate the mass of chromium in grams as follows.
No. of moles = 
mass in grams = No. of moles × Molar mass
= 11.9 moles × 51.99 g/mol
= 618.68 g
Thus, we can conclude that there are 618.68 g in 11.9 moles of chromium.
It indicates that there is only one oxygen molecule
Answer:
The answer to your question is P = 1.64 atm
Explanation:
Data
Volume = 2.5 x 10⁷ L
Temperature = 22°C
Pressure = ?
Moles = 1.7 x 10⁶
R = 0.082 atm L/ mol°K
Process
1.- Convert temperature to °K
Temperature = 22 + 273
= 295°K
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for P
P = nRT / V
- Substitution
P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷
- Simplification
P = 41123000 / 2.5 x 10⁷
- Result
P = 1.64 atm
