Answer:
The effective nuclear charge for a valence electron in oxygen atom: 
Explanation:
Effective nuclear charge
is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.
<em>It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s). </em>

<u>For an oxygen atom</u>-
Electron configuration: (1s²) (2s² 2p⁴)
<em>The atomic number (actual nuclear charge): </em>Z = 8
The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:
⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45
<u><em>Therefore, the effective nuclear charge for a valence electron in oxygen atom is:</em></u>

<u>Therefore, the effective nuclear charge for a valence electron in oxygen atom:</u> 
The answer would be moving bc for friction to happen the fault lines have to be moving
The amu is a unit which means the mass per neutron, So the 74.55 means the mass per mole KCl. So the mole number of 2.330g KCl is 2.330/74.55=0.031 mol. The volume is 0.031/0.1250=0.248 L ≈ 250 mL. So the answer is a.