Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law

Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm (From correct source)
Using above equation as:




The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Mass of the gas m = 1.66
The calculated temperature T = 273 + 20 = 293
We have to calculate molar mass to determine the gas
Molar Mass = mRT / PV
M = (1.66 x 8.314 x 293) / (101.3 x 1000 x 0.001)
M = 4043.76 / 101.3 = 39.92 g/mol
So this gas has to be Argon Ar based on the molar mass.
Pressure is 5.7 atm
<u>Explanation:</u>
P1 = Standard pressure = 1 atm
P2 = ?
V1 = Volume = 10L
V2= 2.4L
T1 = 0°C + 273 K = 273 K
T2 = 100°C + 273 K = 373 K
We have to find the pressure of the gas, by using the gas formula as,

P2 can be found by rewriting the above expression as,

Plugin the above values as,

Answer:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene
Explanation:

Moles of propene = 
According to reaction, 1 mole of propene gives 1 mole of propane.
Then 0.6538 moles of bromo-propane will give:

78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.
<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:

We are given:

Putting values in above equation, we get:

Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 206.7 yrs
= initial amount of the reactant = ?
[A] = amount left after decay process = 1.40 g
Putting values in above equation, we get:
![0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}](https://tex.z-dn.net/?f=0.0101yr%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B206.7yrs%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B1.40%7D)
![[A_o]=11.3g](https://tex.z-dn.net/?f=%5BA_o%5D%3D11.3g)
Hence, the initial amount of Uranium-232 present is 11.3 grams.