No that’s is the electronic configuration for fluorine.
<h3>Given:</h3>
M₁ = 2.0 mol/L
V₁ = 1 L
M₂ = 0.1 mol/L
<h3>Required:</h3>
V₂
<h3>Solution:</h3>
M₁V₁ = M₂V₂
V₂ = M₁V₁ / M₂
V₂ = (2.0 mol/L)(1 L) / (0.1 L)
<u>V₂ = 20 L</u>
Therefore, the volume of the new solution will be 20 L.
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the answer is C.
The PH will initially remain close to 4.76 then decrease quickly when the buffer capacity is exceeded
