Three of them may have decayed more quickly or more slowly than they should have according to the likelihood at that particular moment. However, suppose we have a lot of radioactive new Clyde's, say six times 10 to the 12, and we have three times 10 to the 12 in a minute. The rate may then be averaged out because there are a sufficient number of radioactive new Clyde's. Furthermore, we can say with confidence that the half life is one minute.
<h3>What is radioactivity?</h3>
Radioactivity, as its name suggests, is the act of generating radiation without any external cause. An atomic nucleus that is unstable for whatever reason does this by "wanting" to give up some energy in order to change its configuration to one that is more stable. Modern physics spent a lot of time in the first half of the 20th century figuring out why this occurs, which led to a pretty solid understanding of nuclear decay by 1960. A nucleus with too many neutrons will produce a negative beta particle, which will convert one of the neutrons into a proton. A nucleus with too many protons will emit positrons, which are positively charged electrons that turn protons into neutrons.
To know more about radioactivity:
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The first question would be B) they are examples of cell organelles. The second question would be A) mitochondria. Hope this helps!
Answer:
The second ring in an atom can only hold up to 8 electrons.
The answer to this question would be: alkaline earth metal
Alkali earth metal is the second column group of the periodic table. In this group, the element has 2 extra electrons in their outer cells. That is why most of this metal has 2+ charge.
Their neighbor is the alkali metal which was the first column of the periodic table. The name is similar so don't confused and mix them each other.
Answer:
a. 1810mL
Explanation:
When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:
where the temperatures must be measured in Kelvin
To convert from Celsius to Kelvin, add 273, or use the equation: 
For this problem, one must also recall that standard temperature is 0°C (or 273K).
So,
, and
.

![\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5D%29%7D%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%29%7D)
![\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%29%7D%28322.4%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%20%29%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29%7D%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29)
![1810.04571428[mL]=V_2](https://tex.z-dn.net/?f=1810.04571428%5BmL%5D%3DV_2)
Adjusting for significant figures, this gives ![V_2=1810[mL]](https://tex.z-dn.net/?f=V_2%3D1810%5BmL%5D)