An exponential decay law has the general form: A = Ao * e ^ (-kt) =>
A/Ao = e^(-kt)
Half-life time => A/Ao = 1/2, and t = 4.5 min
=> 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154
Now replace the value of k, Ao = 28g and t = 7 min to find how many grams of Thalium-207 will remain:
A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g
Answer 9.5 g.
Explanation:
To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;
Be = 2, 2
F = 2, 7
Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.
When Be loses two electrons it becomes isoelectronic with He;
Be → Be²⁺ + 2e⁻
Also, when fluorine gains an electron, it becomes isoelectronic with Ne;
F + e⁻ → F⁻
This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.
Hence;
Be²⁺ + 2F⁻ → BeF₂
I'm not so sure but I would say Answer Choice B
The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :
![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)
Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu
Answer:
up down up down thats the pattern
Explanation:
