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yanalaym [24]
3 years ago
13

Write the electronic configurations for mn, mn2+, mn4+, mn6+, and mn7+

Chemistry
1 answer:
mariarad [96]3 years ago
3 0

<u>Answer:</u> The electronic configurations are written below.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.  Number of electrons in an atom is determined by the atomic number of that atom.

Atomic number is defined as the number of protons or electrons that are present in a neutral atom.

Atomic number = number of protons = number of electrons

We know that:

Atomic number of Manganese = 25 = Number of protons

Number of electrons = Number of protons - charge

  • <u>For Mn-atom</u>

Number of electrons = 25

Electronic configuration of Mn : 1s^22s^22p^63s^23p^64s^23d^5

  • <u>For Mn^{2+} ion:</u>

Number of electron = 25 - (+2) = 23

Electronic configuration of Mn^{2+}:1s^22s^22p^63s^23p^63d^5

  • <u>For Mn^{4+} ion:</u>

Number of electron = 25 - (+4) = 21

Electronic configuration of Mn^{4+}:1s^22s^22p^63s^23p^63d^3

  • <u>For Mn^{6+} ion:</u>

Number of electron = 25 - (+6) = 19

Electronic configuration of Mn^{6+}:1s^22s^22p^63s^23p^63d^1

  • <u>For Mn^{7+} ion:</u>

Number of electron = 25 - (+7) = 18

Electronic configuration of Mn^{7+}:1s^22s^22p^63s^23p^6

Hence, the electronic configurations are written above.

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How many grams of ammonia (NH3) can be produced by the synthesis of excess hydrogen gas (H2) and 253.8 grams of nitrogen gas (N2
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Answer:

308.2 g of NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3H₂ + N₂ —> 2NH₃

Next, we shall determine the mass of N₂ that reacted and the mass of NH₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3 = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

Summary:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Finally, we shall determine the mass of NH₃ produced by the reaction of 253.8 g of N₂. This can be obtained as illustrated below:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Therefore, 253.8 g of N₂ will react to produce = (253.8 × 34)/28 = 308.2 g of NH₃.

Thus, 308.2 g of NH₃ were obtained from the reaction.

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