The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g
The equation is Mr=Mi(1/2)^n
where n is the number of half-lives
Mr is the mass remaining after n half lives
Mi is the initial mass of the sample
To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)
n=1000/24100=0.0414
So Mr=5x(1/2)^1=2.5 g
The mass remaining is 2.5 g
- The half life is the time in which the concentration of a substance decreases to half of the initial value.
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Answer:
Option D) Compound B may have a lower molecular weight.
Explanation:
Compound A and B are standing at the same temperature yet compound A is evaporating more slowly than compound B.
This simply indicates that compound B have a lower molecular weight than compound A.
This can further be seen when gasoline and kerosene are placed under same temperature. The gasoline will evaporate faster than kerosene because the molecular weight of the gasoline is low when compared to that of the kerosene.
<span>I think the correct answer is A. A
buffer is a substance that resists small change in the acidity of a solution
when an acid or base is added to the solution. Usually, a buffer involves a
weak acid or a weak alkali and one of its salt.</span>
Answer:
Trial Number of moles
1 0.001249mol
2 0.001232mol
3 0.001187 mol
Explanation:
To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.
Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.
<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.
In this case the solute is <em>NaOH</em>.
The formula is:

Solve for the <em>number of moles:</em>

Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.
Trial mL liters Number of moles
1 12.49 0.01249 0.01249liters × 0.1000M = 0.001249mol
2 12.32 0.01232 0.01232liters × 0.1000M = 0.001232mol
3 11.87 0.01187 0.01187liters × 0.1000M = 0.001187 mol
Answer : The pH of the solution is, 3.41
Explanation :
First we have to calculate the moles of
.


Now we have to calculate the value of
.
The expression used for the calculation of
is,

Now put the value of
in this expression, we get:



The reaction will be:

Initial moles 0.375 0.100 0.375
At eqm. (0.375-0.100) 0 (0.375+0.100)
= 0.275 = 0.475
Now we have to calculate the pH of solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[F^-]}{[HF]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
Now put all the given values in this expression, we get:
![pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]](https://tex.z-dn.net/?f=pH%3D3.17%2B%5Clog%20%5B%5Cfrac%7B%28%5Cfrac%7B0.475%7D%7B1.50%7D%29%7D%7B%28%5Cfrac%7B0.275%7D%7B1.50%7D%29%7D%5D)

Thus, the pH of the solution is, 3.41