`i am holding a balloon containing 439 ml of gas over my fireplace. the temperature and pressure of the gas inside the balloon is 317.15 k and 0.959 atm, respectively. suppose i don't want the pressure to change, but i want to the volume to go down to 0.378 l. what is the temperature that i need to reach when i cool down the balloon? to what temperature (in celsius) must the balloon be cooled to reduce its volume to 0.378 l if the pressure doesn't change (remained constant)?

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law  

Given ,  

V₁ = 439 mL  = 0.439 L ( 1 L = 0.001 mL )

V₂ = 0.378 L

T₁ = 317.15 K

T₂ = ?

Using above equation as:

The conversion of T(K) to T( °C) is shown below:

T( °C) = T(K) - 273.15  

So, T = 273.08 - 273.15 °C = - 0.07 °C

the answer is c i believe

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.