The outer shell electrons are only involved in the bonding process since they are the only 'incomplete' shell and it needs to be fulfilled by another element.
Isn't it a because in b at the start of the equation the E in Fe just disappeared
Answer:
Thus, the radius of the helium atom in nanometers is - 0.031 nm
Explanation:
Given that:-
The radius of the helium atom = 31 pm
Considering the conversion of length in pm to the length in nm as:-
1 pm = 0.001 nm
So,
Applying the above conversion factor in the radius of helium atom as:-
Radius = 
nm = 0.031 nm
<u>Thus, the radius of the helium atom in nanometers is - 0.031 nm</u>
Answer:
a. 0.182
b. 1.009
c. 1.819
Explanation:
Henderson-Hasselbach equation is:
pH = pKa + log [salt / acid]
Let's replace the formula by the given values.
a. 3 = 3.74 + log [salt / acid]
3 - 3.74 = log [salt / acid]
-0.74 = log [salt / acid]
10⁻⁰'⁷⁴ = 0.182
b. 3.744 = 3.74 + log [salt / acid]
3.744 - 3.74 = log [salt / acid]
0.004 = log [salt / acid]
10⁰'⁰⁰⁴ = 1.009
c. 4 = 3.74 + log [salt / acid]
4 - 3.74 = log [salt / acid]
0.26 = log [salt / acid]
10⁰'²⁶ = 1.819
If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation
