In the chamber below, the volume is 20.5 L. What would the volume be if the temperature is increased to 500 K, the moles are reduced to 0.5, and the pressure is increased to 2.0 atm? A.) 10.25 L
B.) 20.5 L
C.) 41.0 L
D.) 61.5 L

C. bc I ssid

Explanation:

0+0+0+0+0+0=00

10.25L

Explanation:

Doubling the temperature would double the volume while halving the number of moles would decrease the volume by half, so the changes produced by these two variables cancel each other out. Since pressure and volume are inversely proportional, doubling the pressure will cut the volume in half, reducing it from 20.5 to 10.25 L.

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.