4 Compared to the energy and charge of the electrons in the first shell of a Be atom, the electrons in the second shell of this atom have(1) less energy and the same charge
(2) less energy and a different charge
(3) more energy and the same charge
(4) more energy and a different charge

Answers

Answer 1
Answer: Moving the electron away from the nucleus requires energy, so the electrons in the outer shell will have more energy than ones in the inner shell. Electrons always have a charge of -1, so the charge in the inner and outer shell will be the same. Therefore the answer is 3
Answer 2
Answer:

The electrons in the second shell of this atom have (3) more energy and the same charge compared to the first shell

Further explanation

Bohr's atomic model has shown the energy levels of electrons in the path of the atomic shell

The greater the value of n (the atomic shell, the main quantum number), the greater the energy level

In normal circumstances, electrons fill the shell at the lowest energy level starting from the shell K, L M and then N

When an atom gets energy from outside, the electrons will absorb energy so it moves to higher energy. This situation is called excited

Electrons will return to the original path or a lower energy level because the excited state is unstable. In this condition, the electron will release energy

The electron energy at the nth path can be formulated:

\rm En=(-Rh)/(n^2)

Rh = constant 2.179.10⁻¹⁸ J

So the electron transfer energy (ΔE)

ΔE = E end - E initial

From the electron transfer available, because the value of the Rh constant is the same, the effect is the value of n (shell) ⇒ 1 / n²

Electron configuration of Be (Beryllium) with atomic number 4

1s² 2s² or [He] 2s²

So in the first shell E = -1(n=1), the second shell E = -1/4(n=2)

So the energy value in the second shell is greater than the first shell

While the electron charge is still the same(-1) (such as protons with + charges and neutrons with neutral charges / 0)

Learn more

statement about electrons and atomic orbitals

brainly.com/question/1832385

Effective nuclear charge

brainly.com/question/5441986

statement about subatomic particles is true

brainly.com/question/3176193


Related Questions

A physician has ordered 0.50 mg of atropine, intramuscularly. If atropine were available as 0.10 mg/mL solution, how many mL would u need to give?
The chemical formula below shows how salt is produced.HCl + NaOH -----> NaCl + H2O Which substances are the reactants?
If one body is positively charged and another body is negatively charged, free electrons tend to A. move from the negatively charged body to the positively charged body. B. remain in the negatively charged body. C. move from the positively charged body to the negatively charged body. D. remain in the positively charged body.
HELPPPPP ASAP PLSSSS !!!!!!!!!
Humans are dependent on _______ for their food supply.a.waterb.biomassc.fossil fuelsd.timber

What is electropositivity and electronegativity? Between group 1 and group 17, which is electropositive and electronegative?

Answers

Electopositivity is the measure of the ability of elements to donate electrons to form positive ions. Whereas the Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. Group 1 is more electropositive than group 17 and group 17 is more electronegative than group 1.

ASAP pleaseee and Thankyou​​

Answers

Answer:

I know the answer, but it is hard to explain

Explanation:

yIt easy u but the single letter in the outside boxes. Then u but the one in the middle box. But the BB in one corner. then but bb on the opposite corner. Then but Bb on the other box.

Answer:

father.....Bb

mother.....bb

offspring, 50%Bb brown eyes ,50%bb blue eyes

State and explain the trends in the atomic radius and ionization energy for elements Li to Cs.

Answers

Explanation:

ionisation energy decrease down the group as the atomic radius increases. Nuclear charge increases. Number of shell increases so the electron will experience more shielding so it would be easier for the atom to loss electron.

Final answer:

The atomic radius increases as you move down a group from Li to Cs, while the atomic radius generally decreases as you move across a period from Li to Cs. The ionization energy decreases down a group and increases across a period.

Explanation:

The atomic radius is the size of an atom, while ionization energy is the energy required to remove an electron from an atom. In general, as we move down a group from Li to Cs, the atomic radius increases due to the addition of more energy levels. This is because the electrons occupy higher energy orbitals farther away from the nucleus. On the other hand, as we move across a period from Li to Cs, the atomic radius generally decreases. This is because the effective nuclear charge increases, pulling the electrons closer to the nucleus.

Regarding ionization energy, it generally decreases down a group from Li to Cs. This is because the atomic radius increases, therefore making it easier to remove an electron from a larger, higher energy orbital. Conversely, as we move across a period, the ionization energy generally increases. This is because the atomic radius decreases, and the electrons are held more tightly by the increasing nuclear charge.

Learn more about Trends in atomic radius and ionization energy here:

brainly.com/question/32859891

#SPJ11

Click on the graphic to choose the correct relationship on the graph.R = diffusion rate of gas
MW = molecular weight

Answers

The graph of the relationship between the rate of diffusion and molar weight is shown by option A

What is the relationship between rate of diffusion and molar weight?

Graham's law of diffusion states that the rate of diffusion is inversely related to the square root of the molar mass of a gas.

According to this relationship, under the same circumstances, lighter molecules disperse more quickly than heavier ones. It is crucial to comprehend how gases behave and how their molecular makeup affects how they travel and are distributed.

Learn more about rate of diffusion:brainly.com/question/30697046

#SPJ3

Answer:

its the first graph

Explanation:

i got the answer correct using odyssey-ware

What did alchemists contribute to the development of chemistry?

Answers

The alchemists were the first chemists, but they were also mixing chemicals, they believed that magic would help with their experiments.but it didn't. but their experiments of the science of chemistry did.

Final answer:

Alchemy made contributions to the development of chemistry through advancements in manipulating matter, discovering the law of conservation of matter, and contributing to the understanding of atoms and the periodic table.

Explanation:

Alchemy made some contributions to the development of chemistry. Although it was not scientific by modern standards, alchemists made progress in manipulating matter and exploring the properties of substances. One example of an alchemist's contribution is the discovery of the law of conservation of matter by Antoine Lavoisier, often regarded as the father of modern chemistry. Lavoisier's work with gases also changed chemistry from a qualitative to a quantitative science and laid the groundwork for further study. Additionally, alchemists played a role in developing the concept of atoms and the periodic table, which are fundamental in chemistry.

Learn more about The contributions of alchemists to the development of chemistry here:

brainly.com/question/4895119

#SPJ6

A 1.0 liter container is filled with 0.300 M ofPCl5 at 250◦C. The vessel is then held at a
constant temperature until the reaction
PCl5(g) ⇀↽ PCl3(g) + Cl2(g)
comes to equilibrium. It is found that the
vessel contains 0.200 moles of PCl5. What is
the value of the equilibrium constant for the
reaction at this temperature?

Answers

If there is 1.0 liters of 0.300 M at the beginning, that means there was 0.300 moles (because 1 L times 0.300 moles/L = 0.300 mol).  It says that, at equilibrium, there is 0.200 mol PCL5.  That means 0.100 mol PCl5 reacted.  Therefore, 0.100 mol each was created of PCl3 and Cl2.

Equilibrium constant Kc can be found with this:
K_c= (concentration\ of\ the\ products)/(concentration\ of\ the\ reactants)\nK_c=([PCl_3][Cl_2])/([PCl_5])=((0.100)(0.100))/((0.200))=0.0500