Answer:
Q = 28.9 kJ
Explanation:
Given that,
Mass of Aluminium, m = 460 g
Initial temperature, 
Final temperature, 
We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

So, 28.9 kJ of heat is required to raise the temperature.
Answer:
The gas that Dr. Brightguy added was O₂
Explanation:
Ideal Gases Law to solve this:
P . V = n . R . T
Firstly, let's convert 736 Torr in atm
736 Torr is atmospheric pressure = 1 atm
20°C = 273 + 20 = 293 T°K
125 mL = 0.125L
0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K
(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n
5.20x10⁻³ mol = n
mass / mol = molar mass
0.1727 g / 5.20x10⁻³ mol = 33.2 g/m
This molar mass corresponds nearly to O₂
One is rows the other is columns
If the half life is 5 days, you would have to divide 222 by 2 3 times.
so 222/2 =111g
111/2 = 55.5g
55.5/2 = 27.75g
after four half life the mass would be
27.75/2
13.875g
the percent would be
13.875/222
= 6.25%