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mylen [45]
3 years ago
11

Ummmm i don’t need help on the question i just need help on how i drag tools with my chrome book i’m stuck i tried everything

Chemistry
1 answer:
Illusion [34]3 years ago
8 0

Answer:

Try using the touch screen and make sure you have not right clicked on anything.

Explanation:

Hope this helps!

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There need be 9 amino acids, as each amino acid is composed of 3 codons.
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Atoms of the same chemical element that have different atomic mass are known as?
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isotopes

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The atoms of a chemical element can exist in different types. These are called isotopes. They have the same number of protons (and electrons), but different numbers of neutrons. Different isotopes of the same element have different masses.

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Energy and Temperature Activity Worksheet
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the ground would heat up slower than the sand because it has lots of moisture  so sand would heat up faster than any thing other than lava

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Rank these transition metal ions in order of decreasing number of unpaired electrons.
lesya692 [45]

Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

  • Atomic number of Fe is 26.

Fe^{3+} - [Ar] 3d^{5}

So, there is only 1 unpaired electron present in Fe^{3+}.

  • Atomic number of Mn is 25.

Mn^{4+} - [Ar]3d^{3}

So, there are only 3 unpaired electrons present in Mn^{4+}.

  • Atomic number of V is 23.

V^{3+} - [Ar] 3d^{2}

So, there are only 2 unpaired electrons present in V^{3+}.

  • Atomic number of Ni is 28.

Ni^{2+} - [Ar] 3d^{8}

So, there will be 2 unpaired electrons present in Ni^{2+}.

  • Atomic number of Cu is 29.

Cu^{+} - [Ar] 3d^{10}

So, there is no unpaired electron present in Cu^{+}.

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

7 0
3 years ago
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