They drill too deep and find lava
Answer:
46.8% should be. correct, I'm not fully sure though. : )
<u>Answer: </u>
A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed
<u>Explanation:</u>
Given, the initial value of the sample,
= 150mg
Final value of the sample or the quantity left, A = 18.75mg
Time = 11.4 days
The amount left after first half life will be ½.
The number of half-life is calculated by the formula

where N is the no. of half life
Substituting the values,


On equating, we get, N = 3
Therefore, 3 half-lives have passed.
Answer:
c. ΔH° is positive and ΔS° is positive.
Explanation:
Hello,
In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:

Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.
Best regards.
Of Uranium-235 remains after 2.8 x 10^9 years, what was the original mass of the sample of Uranium-235? The half-life of Uranium-235 is 7.0 x 10^8 years. Uranium-232 has a half life of 68.8 years.