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TiliK225 [7]
3 years ago
10

1) According to Bohr's model of the atom, in which orbitals do electrons have the most energy?

Chemistry
2 answers:
Alex Ar [27]3 years ago
7 0

The orbitals closest to the nucleus is the orbital wih the lowest energy. This is according to the basic rules stating that the energy of the shells as its principal quantum number increases, also increases. Thus the answer in 1 is B. Valence electrons are found in the outermost electron shell, on the other hand. 
FinnZ [79.3K]3 years ago
6 0

The answer is:

1) Outermost orbitals have the most energy.

The explanation:

In Bohr model we can see that the highest energy level is found in the outermost orbital , when the first orbital form and make a shell around the nucleus and its principal quantum number which has the symbol (n) is = 1

So the other orbital away from the nucleus is gradually assigned values:

n = 2  

n= 3

n=4

etc

and  the most stable and the lowest energy level is found in the innermost orbit.  

2) valence electrons located in the outermost electrons shell.

The explanation:

when:  

The number of electrons in the outermost shell of an atom gets its reactivity or tendency from it to form and make the chemical bonds with the other atoms.

This outermost shell is called the valence shell, and the electrons found in this shell are called valence electrons.

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The Earth's gravitational force accelerates objects when they fall. It constantly pulls, and the objects constantly speed up.
8 0
2 years ago
If an element has an atomic number of 20 and a mass number of 41, what is the :
iVinArrow [24]

Answer:

20 protons, 20 electrons, and 21 neutrons

Explanation:

The atomic number of an atom is the number of protons it has. If the atomic number is 20 then we know the atom has 20 protons.

•The mass number of an atom is the total number of protons and neutrons the atom contains. The mass number is 41 and the number of protons is 20, just subtract 20 from 41 and you will get the number of neutrons: 41 - 20= 21. The atoms has 21 neutrons.

•The number of electrons found in an atom is equal to the number of protons. The atoms has 20 protons which means it has 20 electrons.

So, the answer is:

20 protons, 20 electrons, and 21 neutrons

8 0
3 years ago
Which element in Group 17 is the most active nonmetal? (1) Br (2) I (3) Cl (4) F
Maksim231197 [3]

The element of the group 17 that is most active non metal is fluorine.

The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).

Among all the elements of the group 17. Fluorine is the smallest in size.

Because of the small size of fluorine it has the highest electronegativity in group 17.

This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.

Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.

It reacts readily to form oxides and hydroxides.

So, we can conclude here that fluorine is the most active non metal of group 17.

To know more about group 17, visit,

brainly.com/question/26440054

#SPJ4

7 0
1 year ago
Classify each change as physical or chemical.
trasher [3.6K]

A chemical

B physical

C physical

D Chemical

4 0
3 years ago
Which of these half-reactions represents reduction?
gogolik [260]

Answer: The half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in Cr_{2}O^{2-}_{7} is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}

Similarly, oxidation state of Mn in MnO^{-}_{4} is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}

Thus, we can conclude that half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}
3 0
3 years ago
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