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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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Answer:

1.73g of CO2.

Explanation:

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

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Next we shall determine the masses of NaHCO3 and CH3COOH that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 84g/mol

Mass of NaHCO3 from the balanced equation = 1 x 84 = 84g

Molar mass of CH3COOH = 12 + (3x1) + 12 + 16 + 16 + 1 = 60g/mol

Mass of CH3COOH from the balanced equation = 1 x 60 = 60g

Molar mass of CO3 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH to produce 44g of CO2.

Next, we shall determine the limiting reactant of the reaction. This is illustrated below:

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH.

Therefore, 3.3g of NaHCO3 will react with = (3.3 x 60)/84 = 2.36g of CH3COOH.

From the above illustration, we can see that only 2.36g of CH3COOH out of 10.3g given reacted completely with 3.3g of NaHCO3. Therefore, NaHCO3 is the limiting reactant while CH3COOH is the excess reactant.

Finally, can determine the mass of CO2 produced during the reaction.

In this case the limiting reactant will be used because it will produce the mass yield of CO2 as all of it were used up in the reaction. The limiting reactant is NaHCO3 and the mass of CO2 produced is obtained as shown below:

From the balanced equation above,

84g of NaHCO3 reacted to produce 44g of CO2.

Therefore, 3.3g of NaHCO3 will react to produce = (3.3 x 44)/84 = 1.73g of CO2.

Therefore, 1.73g of CO2 is released during the reaction.

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The vapor pressure of pure pentane and pure hexane are 425 torr and 151 torr respectively. equal moles of pentane and hexane are
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The mole fraction of pentane in the vapor is 0.291

Vapour pressure rises with temperature and is a measurement of a substance's propensity to transform into a gaseous or vapour state. The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapour pressure present at the liquid's surface.

The number of moles of a particular component in the solution divided by the overall number of moles in the sample solution is known as the mole fraction.

Using the formula for vapour pressure,

vapour pressure = P^{}°_{hex} × X_{hex} + P^{}°_{pen} × X_{pen}

vapour pressure = 151 ×  (1-X_{pen}) + 425 × X_{pen}

240 = 151 - 151X_{pen} + 425X_{pen}

240 - 151 = -  151X_{pen} + 425 X_{pen}

89 = 274 X_{pen}

\frac{89}{274} = X_{pen}

0.291 = X_{pen}

Therefore,  the mole fraction of pentane in the vapor is 0.291

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