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Luden [163]
3 years ago
10

Which of the following is the correct rank of increasing electron affinities (from less exothermic to more exothermic)? Select t

he correct answer below: Cs
Chemistry
1 answer:
Lena [83]3 years ago
8 0

The question is incomplete, the complete question is ;

Which of the following is the correct rank of increasing electron affinities (from less exothermic to more exothermic)? Select the correct answer below: O Cs <K<Na <Li O Li< Na <K <Cs O Na <Li<K <Cs OK<Cs < Li< Na

Answer:

Cs< K<Na<Li

Explanation:

The electron affinity is defined as the energy released when a neutral gaseous atom accepts an electron to create a gaseous negative ion. This reaction is exothermic and leads to the evolution of heat.

When an electron is added to a metal, large energy is required since metals are usually electro positive. Electron affinity decreases down the group since electrons enter into higher energy levels further away from the nucleus. They experience lesser nuclear attraction.

This is clearly evidenced in the sharp decrease in the electron affinity of group one elements from lithium to caesium. The electron affinity of lithium is 60Kjmol-1 but decreases steadily down the group to a value of 46KJmol-1 for caesium. This explains the order chosen in the answer.

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

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

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

Putting the values into the above formula as follows.

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

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, 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

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

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

Hence, before the reaction:

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

On the other hand, 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) mol

                                = 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.

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

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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