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.
Answer:
The answer is

Explanation:
The energy of a quantum of light can be found by using the formula
<h3>E = hf</h3>
where
E is the energy
f is the frequency
h is the Planck's constant which is
6.626 × 10-³⁴ Js
From the question
f = 4.31 × 10¹⁴ Hz
We have
E = 4.31 × 10¹⁴ × 6.626 × 10-³⁴
We have the final answer as

Hope this helps you
Answer:
M=0.15
Explanation:
138 g AgNO -> 1 mol AgNO
10 g AgNO -> x
x= (10 g AgNO * 1 mol AgNO)/138 g x=0.07 mol AgNO
450 mL=0.45 L
M= mol solute/L solution
M= 0.07 mol AgNO/0.45L
M=0.15
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