Hope this helps! If you dont understand balancing equations in general, say so in the comments, I’m happy to help
<span>The composition of a fertilizer is usually express in NPK number. NPK number is in terms of Percent by mass of the said element which are Nitrogen, Phosphorus and Potassium. A 15-35-15 fertilizer has 15%
Nitrogen, 35% Phosphorous, and 15% Potassium by mass. If you have 10 g of this
fertilizer, to get the number of moles of phosphorus, you multiply the mass by
35%, which is equal to 10*0.35 or 3.5 g phosphorus. Then you divide the
calculated mass of phosphorous by its molar mass which is 30.97 g/mol.
Therefore, you have 3.5/30.97 which is equal to 0.1130 mol Phosphorus. This is the amount of Phosphorus in moles in the fertilizer.</span>
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.
More precisely, we need to specify its position<span> relative to a convenient reference frame. .... Also you s</span>hould know<span> that some people use the subscript "0" to refer to the ... mx, </span>start<span> subscript, 0, end subscript, equals, 1, </span>point<span>, 5, space, m and her </span>final<span> ... </span>between<span> two </span>points<span>, or we </span>can<span> talk about the distance traveled by an </span>object<span>.</span>
