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Butoxors [25]
3 years ago
15

Suppose a new group 5A element is discovered. Photoelectron spectroscopy (PES) reveals that its last two peaks occur at 1370 kJ/

mol and 777 kJ/mol. Calculate the average valence electron energy (AVEE) of this element.
Chemistry
1 answer:
iren2701 [21]3 years ago
5 0

Answer: The average valence electron energy (AVEE) of this element =

1014.2 KJ/ mol  or  1.0142mJ/mol.

Explanation:

The average valence electron energy =  (number of electrons in s subshell  x  Ionization energy of that subshell) + (number of electrons in p subshell  x  Ionization energy of that subshell) / total number of electrons in both subshells of the valence shells.

The 5A elements are non-metals like Nitrogen and Phosphorus with the metallic character increasing as you go down the group, So a new 5A element will have characteristics of its group with 5 valence electron in its outermost shell represented as ns2 np3

Therefore  the average valence electron energy (AVEE) of this element will be calculated as

The average valence electron energy = (2 x 1370  kJ/mol + 3 x 777 kJ/mol.) / 5

2740+2331/ 5 =5071/5

=1014.2 KJ/ mol  or  1.0142mJ/mol.

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Answer:

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Excerpt from textbook

8 0
3 years ago
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inessss [21]

Answer:

6.

a. D

b. F

Explanation:

4 0
3 years ago
How do chemists solve problems?
Westkost [7]

Answer:Chemistry problems can be solved using a variety of techniques.

Explanation:  Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!

4 0
3 years ago
The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of
Olin [163]

Answer:

            Empirical Formula  =  C₃H₆O₁

Solution:

Data Given:

                      Mass of Ethyl Butyrate  =  3.61 mg  =  0.00361 g

                      Mass of CO₂  =  8.22 mg  =  0.00822 g

                      Mass of H₂O  =  3.35 mg  =  0.00335 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100

                      %C  =  (2.277) × (12 ÷ 44) × 100

                      %C  =  2.277 × 0.2727 × 100

                      %C  =  62.09 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.9279) × (2.02 ÷ 18.02) × 100

                      %H  =  0.9279 × 0.1120 × 100

                     %H  =  10.39 %


                      %O  =  100% - (%C + %H)

                      %O  =  100% - (62.09% + 10.39%)

                      %O  =  100% - 72.48%

                      %O  =  27.52 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 62.09 ÷ 12.01

                      Moles of C  =  5.169 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 10.39 ÷ 1.01

                      Moles of H  =  10.287 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 27.52 ÷ 16.0

                     Moles of O  =  1.720 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

             5.169                                10.287                              1.720

       5.169/1.720                       10.287/1.720                     1.720/1.720

               3.00                                   5.98                                   1

                  3                                      ≈ 6                                     1

Result:

         Empirical Formula  =  C₃H₆O₁

8 0
3 years ago
What is Kb for CH3NH2(aq) + H2O(1) CH3NH3(aq) + OH (aq)?
statuscvo [17]

Answer:

Kb for CH₃NH₂ (methylamine) is 4.4 × 10⁻⁴

Hope that helps.

8 0
3 years ago
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