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3 years ago

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Arrange the following sublevels in order of increasing energy: 3d, 2s, 4s, 3p.

1 answer:

ycow [4]3 years ago

6 0

To arrange the sublevels in order of increasing energy, we add the principal shell and the maximum number of electrons of the specific sublevel. s has 2 electrons, p has 6 electrons, d has 10 electrons, and f and 14 electrons. in this case, a) 3 + 10 = 13b) 2 + 2 = 4c) 4 + 2 = 6d) 3 + 6 = 9
The order hence is 2s, 4s, 3p, and 3d

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Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

Moles of acetic acid = 0.0100 M × 1.00 L

= 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

= 0.100 mol

When $HNO_{3}$ is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles $HNO_{3}$ )

= 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles $HNO_{3}$ )

= 0.100 mol - x

According to Henderson - Hasselbalch equation,

pH = $pK_{a} + log\frac{[conjugate base]}{[weak acid]}$

pH = $pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}$

4.95 = 4.75 + $log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$

$log\frac{(0.100 mol - x)}{(0.0100 mol + x)}$ = 4.95 - 4.75

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= 1.6

Hence, x = 0.032555 mol

Therefore, moles of $HNO_{3}$ = 0.032555 mol

volume of $HNO_{3}$ = $\frac{moles HNO_{3}}{molarity of HNO_{3}}$

= $\frac{0.032555 mol}{10.0 M}$

= 0.0032555 L

or, = 3.25 (as 1 L = 1000 mL)

Thus, we can conclude that volume of $HNO_{3}$ added is 3.26 mL.

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