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Schach [20]
4 years ago
12

How many electrons in an atom can share the quantum numbers n = 1, m subscript l = 0? A. 1 B.2 C.3 D.6

Chemistry
1 answer:
Kisachek [45]4 years ago
6 0
I believe it would be c: 3 because you can have the numbers -1,0,1.
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A buffer consists of 0.120 m hno2 and 0.150 m nano2 at 25°c.
wolverine [178]

Answer:

A) pH of the buffer is 3.44

B) pH of the buffer solution is 3.37

Explanation:

Relation between K_{a} and pK_{a} is as follows.

pK_{a} = -log (K_{a})

. pK_{a} = -log (K_{a})                                    = -log(4.5 \times 10^{-4})                                     = 3.347

The relation between pH and  pK_{a} is as follows.

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

= 3.347+ log \frac{0.15}{0.12}                    = 3.44

pH of the buffer is 3.44.

b)

mol of HCl added = 11.6M *0.001 L = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to formHNO_{2}

No. of moles of

NO^{-}_{2} = 0.15 M \times 1.0 L                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                              = 0.12 mol

after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                         = (0.15 - 0.0116) mol

                                         = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                = (0.12 + 0.0116)

                = 0.1316 mol

As,

K_{a} = 4.5 \times 10^{-4}           pK_{a} = -log (K_{a})                         = -log(4.5 \times 10^{-4})                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

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

= 3.347+ log {0.1384/0.1316}

           = 3.369

           ≅ 3.37

pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37

4 0
3 years ago
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