Answer:
2.15 mg of uranium-238 decays
Explanation:
For decay of radioactive nuclide-
![N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}](https://tex.z-dn.net/?f=N%3DN_%7B0%7D.%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7Bt%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7D)
where N is amount of radioactive nuclide after t time,
is initial amount of radioactive nuclide and
is half life of radioactive nuclide
Here
,
and ![t_{\frac{1}{2}}=4.5\times 10^{9}yr](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D4.5%5Ctimes%2010%5E%7B9%7Dyr)
So,![N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}](https://tex.z-dn.net/?f=N%3D%284.60mg%29%5Ctimes%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B%5Cfrac%7B4.1%5Ctimes%2010%5E%7B9%7D%7D%7B4.5%5Ctimes%2010%5E%7B9%7D%7D%7D)
so, N = 2.446 mg
mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg
In group theory, a branch of mathematics, the term order is used in two unrelated senses:
<span><span>The order of a group is its cardinality, i.e., the number of elements in its set. Also, the order, sometimes period, of an element a of a group is the smallest positive integer m such that <span>am = e</span> (where e denotes the identity element of the group, and am denotes the product of m copies of a). If no such m exists, a is said to have infinite order.</span><span>The ordering relation of a partially or totally ordered group.</span></span>
This article is about the first sense of order.
The order of a group G is denoted by ord(G) or | G | and the order of an element a is denoted by ord(a) or | a |.
Answer:
"1 M" will be the right solution.
Explanation:
The given values are:
Number of moles,
= 2 moles
Volume of solution,
= 2000 mL
or,
= 2 L
Now,
The molarity of the solution will be:
= ![\frac{Moles}{Liters}](https://tex.z-dn.net/?f=%5Cfrac%7BMoles%7D%7BLiters%7D)
On substituting the values, we get
= ![\frac{2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B2%7D)
= ![1 \ M](https://tex.z-dn.net/?f=1%20%5C%20M)
Answer : The time taken by the reaction is ![2.2\times 10^2s](https://tex.z-dn.net/?f=2.2%5Ctimes%2010%5E2s)
Explanation :
The expression used for second order kinetics is:
![kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}](https://tex.z-dn.net/?f=kt%3D%5Cfrac%7B1%7D%7B%5BA_t%5D%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
where,
k = rate constant = ![9.4\times 10^{-3}M^{-1}s^{-1}](https://tex.z-dn.net/?f=9.4%5Ctimes%2010%5E%7B-3%7DM%5E%7B-1%7Ds%5E%7B-1%7D)
t = time = ?
= final concentration = 0.40 M
= initial concentration = 2.16 M
Now put all the given values in the above expression, we get:
![(9.4\times 10^{-3})\times t=\frac{1}{0.40}-\frac{1}{2.16}](https://tex.z-dn.net/?f=%289.4%5Ctimes%2010%5E%7B-3%7D%29%5Ctimes%20t%3D%5Cfrac%7B1%7D%7B0.40%7D-%5Cfrac%7B1%7D%7B2.16%7D)
![t=216.706s\aprrox 2.2\times 10^2s](https://tex.z-dn.net/?f=t%3D216.706s%5Caprrox%202.2%5Ctimes%2010%5E2s)
Therefore, the time taken by the reaction is ![2.2\times 10^2s](https://tex.z-dn.net/?f=2.2%5Ctimes%2010%5E2s)