<h3>You will pay $ 30876800</h3>
We'll begin by calculating the mass in ounce (oz) of a cube foot (ft³) of gold. This can be obtained as follow:
<h3 />
Density of gold = 19298 oz/ft³
Volume of gold = 1 ft³
<h3>Mass of gold =?</h3>
Density = mass /volume
19298 = mass / 1
<h3>Mass of gold = 19298 oz</h3>
Finally, we shall determine the cost of 19298 oz of gold. This can be obtained as follow:
1 oz = $ 1600
Therefore,
19298 oz = 19298 × 1600
19298 oz = $ 30876800
Therefore, a solid cube foot of gold (i.e 19298 oz) will cost $ 30876800
Learn more: brainly.com/question/15407624
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
Because the oily part of its molecule tends to lay up against solid surfaces, but without much in the way of areas of electric charge that would cause it to stick more firmly to that surface.
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Answer:
Rn
Explanation:
Rn is the symbol for radon with the atomic number of 86