Answer:
mook passes between the earth and sun
Answer: Option (D) is the correct answer.
Explanation:
Valence shell is the shell present on the outermost core of an atom and electrons present in the valence shell are known as valence electrons.
If an atom has completely filled valence shell then it means the atom is not reactive in nature because it is already stable.
But when an atom has less than eight electrons in its valence shell then it means to attain stability the atom will readily attract electrons towards itself.
As the given element 1 has 8 electrons in its valence shell. Hence, it is not reactive in nature but element 2 has 6 valence electrons. So, in order to attain stability element 2 will readily attract 2 electrons from a donor atom.
Thus, we can conclude that element 2 is more reactive because it does not have a full valence shell, so it will attract electrons.
I think it will be letter A tbh
Hope this helps you out
<u>Answer:</u>
<u>Balancin</u>g is making the number of atoms of each element same on both the sides (reactant and product side).
To find the number of atoms of each element we multiply coefficient and the subscript
For example contains
5 × 1 = 5,Ca atoms and
5 × 2 = 10, Cl atoms
If there is a bracket in the chemical formula
For example we multiply coefficient subscript number outside the bracket.......... to find the number of atoms
(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )
So
3 × 3 = 9, Ca atoms
3 × 1 × 2 = 6, P atoms
3 × 4 × 2 = 24, O atoms are present.
So
Let us balance the equation given
(Unbalanced)
Reactant side - Number of atoms of each element - Product side
1 - Sb - 1
2 - Cl - 3
So we have on the product side odd number of Cl to convert to even number, multiply by 2
The equation changes to
Reactant side - Number of atoms of each element - Product side
1 - Sb - 2
2 - Cl - 6
Multiplying Sb by 2 and
The equation is balanced now!!!!
Reactant side - Number of atoms of each element - Product side
2 - Sb - 2
6 - Cl - 6
Balanced.