Correct Answer: The Sun heats the Earth unevenly; this heating pattern then causes convection currents in the atmosphere.
There are several information's already given in the question. Based on those information's, the answer can be easily deduced.
Amount of gasoline required by Harry's car to travel 25 miles = 1 gallon
Then
amount of gasoline required
by Harry's car to travel 15000 miles = 15000/25
= 600 gallons
So
Amount of CO2 released by burning 1 gallon of gasoline = 20 pounds
Then
Amount of CO2 released
by burning 600 gallon of gasoline = 600 * 20
= 12000 pounds
From the above deduction, it can be concluded that the amount of CO2 that will be added by Harry's car to the atmosphere is 12000 pounds.
Answer: <span>A-Ce is oxidized because it is losing electrons and Cu is reduced because it is gaining electrons</span><span>.
</span>There are two reactions in the equation, oxidation and reduction. A molecule that oxidized will lose electrons while the molecule that reduced will gain electrons. In this case, Cu2+ changed into Cu which means its oxidation number reduced from +2 into 0. Ce oxidation number increased from 0 into +3
Answer:

Explanation:
You look at the type of atom and their electronegativity difference.
If ΔEN <1.6, covalent; if ΔEN >1.6, ionic
Ar/Xe: Noble gases; no reaction
F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic
N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.
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