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Stells [14]
3 years ago
8

Compare the three atom diagrams. which one shows the electron with the highest potential energy

Chemistry
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer : The one diagram which shows the electron with the highest potential energy is attached below.

Explanation : One can easily find the highest potential energy of the atom just by looking at the diagram, the electron which is from farthest distance from the atomic nucleus will have the highest potential energy in the electron.

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At 50 degrees Celsius and standard pressure inter-molecular forces of attraction are strongest in a sample of ethanoic acid. 
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Why are valence electrons important?
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They are the outer layer of the electron layers.
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Which of the following is/are correct regarding the pentose phosphate pathway? I. NADH is generated by the oxidation of glucose-
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Answer:

The correct answer is 1 NADH is generated by the oxidation of glucose-6-phosphate.

Explanation:

Pentose phosphate pathway deals with the utilization of glucose-6-phosphate by oxidation process to form 6-phosphogluconolactone by the catalytic activity of glucose-6-phosphate dehydrogenese.

   This enzyme need NAD+ as co enzyme which get reduced to generate NADH.

4 0
3 years ago
What is the ratio of effusion rates for the lightest gas, h2, to the heaviest known gas, uf6?
andriy [413]

The ratio of effusion rates for the lightest gas H₂ to the heaviest known gas UF₆ is 13.21 to 1

<h3>What is effusion?</h3>

Effusion is a process by which a gas escapes from its container through a tiny hole into evacuated space.

Rate of effusion ∝ 1/√Ц, (where Ц is molar mass)

Rate H₂ = 1/√ЦH₂

Rate UF₆ = 1/√ЦUF₆

Therefore, Rate H₂/ Rate UF₆ = √ЦH₂/√ЦUF₆

ЦH₂= 2.016 g/mol

ЦUF₆= 352.04 g/mol

Rate H₂ / Rate UF₆ = √352.04/√2.016 = 18.76/1.42

Rate H₂ / Rate UF₆ = 13.21

Therefore, H₂ is lower mass than UF₆. Thus H₂ gas will effuse 13 times more faster than UF₆ because the most probable speed of H₂ molecule is higher; therefore, more molecules escapes per unit time.

learn more about effusion rate: brainly.com/question/28371955

#SPJ1

8 0
1 year ago
What is the molarity of a solution prepared by dissolving 30.0g NaCl in enough water to make a 0.300 L solution?
meriva

Answer:

D.) 1.71 M NaCl

Explanation:

Molarity equation: M= n/v

n= moles of solute

v=liters of solution

NaCl= 58.443 g/mol

30g NaCl / 58.443g/mol = 0.5133(this is n)

0.5133 mols/0.300 L=1.71115674 M

5 0
3 years ago
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