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3 years ago

How many outer-shell d electrons are there in each of the following transition-metal ions?

Chemistry

1 answer:

lesya692 [45]3 years ago

4 0

Answer:

The answer to your question is:

Re (III) has 5 electrons

Sc(III) = has 1 electron

Ru(IV) = has 6 electrons

Hg(II) = has 10 electrons

Explanation:

75 Re(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁵

21 Sc(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹

44 Ru(IV) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁶

80 Hg(II) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰

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In order to complete his research project, Roger needs to make a mixture of 86 mL of a 36% acid solution from a 39% acid

Zina [86]

The volume of the 39% acid solution that Roger needs to use to make a mixture of 86 mL of a 36% acid solution is 74.27 mL.

The mixture of the acids solutions is given by:

$CV = C_{1}V_{1} + C_{2}V_{2}$ (1)

Where:

C: is the concentration if the mixture = 36%

V: is the total volume of the mixture = 86 mL

C₁: is the concentration of acid 1 = 39%

V₁: is the volume if acid 1 =?

C₂: is the concentration of acid 2 = 17%

V₂: is the volume of acid 2

The sum of V₁ and V₂ must be equal to V, so:

$V = V_{1} + V_{2}$

$V_{2} = V - V_{1}$ (2)

By entering equation (2) into (1), we have:

$CV = C_{1}V_{1} + C_{2}(V - V_{1})$

$36\%*86 mL = 39\%*V_{1} + 17\%(86 mL - V_{1})$

Changing the percent values to decimal ones:

$0.36*86 mL = 0.39*V_{1} + 0.17(86 mL - V_{1})$

Now, by solving the above equation for V₁:

$V_{1} = 74.27 mL$

Therefore, the volume of the 39% acid solution is 74.27 mL.

To learn more about mixture and solutions, go here: brainly.com/question/6358654?referrer=searchResults

I hope it helps you!

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Finger [1]

Answer:

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Explanation:

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andreev551 [17]

In the choices above, the correct letter of the question is letter a, scintillation counter. It is because it uses a phosphor-coated surface in order to use its function in detecting radiation. Its component could produce bright flashes where it will be sufficient in detecting the radiation.

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3 years ago

The electron configuration that belongs to the atom with the lowest first ionization energy is?

sasho [114]

The electron configuration that belongs to the atom with the lowest first ionization energy is francium.

<h3>What is ionization energy? </h3>

Ionization energy is defined as the minimum amount of energy required to remove the most loosely electron present in outermost shell.

<h3>Ionization energy across period</h3>

Ionization energy increase as we move from left to right in the period. This can be explained as when we move from left to right along period new electron is added to the same shell which increase the nuclear charge. Hence results int he decrease in size. Due to this decrease in size more energy is required to remove electron from outermost shell.

<h3>Ionization energy along group</h3>

Ionization energy decrease as we move from top to bottom along group. This can be explained as we move from top to bottom new electron is added to new shell. Due to addition of new shell the size of atom increases which results in the decrease in the nuclear charge. Due to this less amount of energy is needed to remove an electron.

Thus, we concluded that the electron configuration that belongs to the atom with the lowest first ionization energy is francium.

learn more about ionization energy:

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