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noname [10]
2 years ago
6

Why chlorine has high electroaffinity than flourine​

Chemistry
1 answer:
Aleksandr-060686 [28]2 years ago
7 0

Answer:

Fluorine has seven electrons in 2p-subshell whereas chlorine has seven electrons in its 3p-subshell. 3p-subshell is relatively larger than 2p-subshell. Therefore, repulsion among the electrons will be more in the 2p-shell of fluorine than 3p-subshell in chlorine.  Due to the smaller size and thus, the greater electron-electron repulsions, fluorine will not accept an incoming electron with the same as chlorine.

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The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (n_{A}) = 1

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Molarity of the base, NaOH (M_{B}) = 0.505 M

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<h3>Volume of the base, NaOH (V_{B}) =? </h3>

\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}

Cross multiply

0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4

Divide both side by 0.505

V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

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