Here’s the math for your answer, which is 3.3 L HCl
Answer: |x+2|=xF2+4
Explanation: not sure if this is right im just guessing
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:


Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 =
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)


0.9715 Fraction of Pu-239 will be remain after 1000 years.
Q = mct
-Q= energy in Joules
-m = mass in grams
-c= specific heat capacity in J/g degree C
-t = delta temperature in degrees Celsius
So,
Q = m c t
Q = (7 grams)(0.448J/g C)(750 C - 25 C)
Q = 2273.6 J
Your final answer = 2273.6 Joules