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irina [24]
3 years ago
13

elements on the right side of the periodic table differ from the elements on the left side in that elements on the right side

Chemistry
1 answer:
yarga [219]3 years ago
7 0

Element  on the right side  of the periodic  table  differ from the elements on  the left side  in that elements on the <em>right  side  are non   metallic  and tends to be gases at room  temperature.</em>


<em>  </em><u>Explanation</u>

In the periodic table there element in  the right side , left side and those  which are in between.

  •  Example of element  in the right side is  fluorine  chlorine, neon, Argon   among others.
  • This element have higher  effective nuclear  charges   and  stabilize electrons more effectively.
  • there electrostatic  intermolecular  forces  are generally weak  therefore they  exist in  liquid or gaseous state.
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Answer:

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Nitrogen is an atom made up of 7 electrons.

To draw the orbital energy level diagram, let us write the orbital notation of the atom;

  7 electrons of Nitrogen:

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So,

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            1s²         2s²          2p³

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When molten rock undergoes crystallization above ground, what type of rock results?
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An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t
Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

5 0
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