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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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3 years ago
The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0
3 years ago
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