What is the pressure of 0.540 mol of an ideal gas at 35,5 l and 223 k?
use pv = nrt and r= 8.314 okpa
molok
0.715 kpa
2.45 kpa
28.2 kpa
62.7 kpa
mark this and return

P = 28.2 Kpa

Explanation:

Given data:

Volume = 35.5 L

Temperature = 223 K

Number of moles = 0.540 mol

R = 8.314 Kpa. L/mol.K

Pressure = ?

Solution:

PV = nRT

P = nRT / V

P = 0.540 mol . 8.314 Kpa. L/mol.K .223 K / 35.5 L

P = 1001.2 Kpa . L /35.5 L

P = 28.2 Kpa

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.