2 im pretty sure but im not for sure
Where's the list? It depends on which one it is. It could be Chlorine, silicon, or anything on the non metals side of the periodic table if you don't list it.
According to that each half-life we lose half of the concentration of N2O so,
Because we start with a concentration of N2O = 0.25 mol
so after one half-life the concentration of N2O decrease to the half 0.25/2
= 12.5 x 10^-2 M
and after two half-lives the concentration of N2O of the one half-life decrease to the half (12.5 x 10^-2) / 2
=6.25 x 10^-2 M
and after three half-lives the concentration of N2O of the two half-lives decrease to the half (6.25x10^-2) / 2
= 3.1 x 10^-2 M
∴ your correct answer is 3.1 x 10^-2 M
Answer:
1.33 kilograms of this liquid must be evaporated to freeze a tray of water at 0°C to ice at 0°C.
Explanation:
Let the mass of liquefied dichlorodifluoromethane be x.
Mass of water to freeze = 571 g
Moles of water =
Heat of fusion of ice = 6.02 kJ/mol
Heat lost when 1 mole of water freeze's = -6.02kJ/mol
Heat lost when 31.7 moles of water freeze's: Q

Heat required to evaporate x amount of liquefied dichlorodifluoromethane: Q'
Q'= -(Q) = 191. kJ
Moles of liquefied dichlorodifluoromethane =
Heat of vaporization of dichlorodifluoromethane = 17.4 kJ/mol


Solving for x:
x = 1328.2 g = 1.33 kg
1 g = 0.001 kg
1.33 kilograms of this liquid must be evaporated to freeze a tray of water at 0°C to ice at 0°C.