Forest because there is increased vegetation that would burn
Answer: with the atomic number.
Explanation:
The table is missing, but I can explain you how to do it, which is the best thing for you because it will permit you understand and solve problems of this kind by yourself.
1) Every element in the periodic table (there are 118 element) are ordered by atomic number. Not two elements have the same atomic number and all the elements from tomic number 1 to atomic number 118 are listed on the periodic table.
2) The atomic number is the number of protons in the nucleus of the element.
3) The atoms of the elements are neutral particles, and given that the magnitude of the charge of a proton is the same than the magnitude of the charge of an electron, the number of electons in an atom is the same number of protons.
4) Here are some examples:
Element/symbol atomic number number of protons number of electrons
Hydrogen, H 1 1 1
Helium, He 2 2 2
Lithium, Li 3 3 3
Berilium. Be 4 4 4
Boron, B 5 5 5
Carbon, C 6 6 6
For ions it is different, because ions are atoms with charges, this is they have lost or gained electrons.
Every positive charge on an ion means that it has loss one electron.
Every negative charge on an ion means that it has gain one electron.
Answer:
A: Outer planet (most likely Jupiter)
B: Inner planet (Mars)
C: Outer planet (maybe Uranus)
D: Outer planet (possibly Saturn)
E: Inner (most likely Mercury)
F: Inner (may be Venus)
Of the same kind or type — homogeneous
The amount of water that falls on the earth — precipitation
Area having a dominant or unifying characteristic — region
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Answer:
Atmospheric density,
Given:
Gas constant for dry air, R = 287
Temperature of gas,
Pressure of gas,
Solution:
Now, to calculate the atmospheric density, we follow:
The eqn for an ideal gas is given by:
where
n = no. of moles of gas
n =
Also, we can write,
Comparing both the values of n, we get:
(1)
Now, the molecular mass of air can be calculated as:
(Since, Oxygen and nitrogen constitutes to about 21% and 78% in the atmosphere)
Using the value above in eqn (1):