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sertanlavr [38]
2 years ago
14

(5 points)

Chemistry
1 answer:
ExtremeBDS [4]2 years ago
5 0

<u>Answer 2 :</u> The given electronic configuration for a neutral atom of phosphorous in its ground state is incorrect.

Explanation :

A neutral atom of phosphorous has 15 electrons.

The given electronic configuration is incorrect.

The reason is, According to Aufbau principle, the electrons will be first filled in the sub-shell having lower orbital energy. As from the given configuration, 3p sub-shell has lower orbital energy than 4s sub-shell. So, the electrons will be filled in 3p sub-shell first. Hence, the ground state electronic configuration of neutral atom of phosphorous is,

1s^22s^22p^63s^23p^3

<u>Answer 3 :</u>

Element                     Rubidium              Magnesium                Aluminium

Symbol                             Rb                         Mg                              Al

Group number                  1                             2                               13

Number of valence          1                             2                                3

electrons

The order of general reactivity on the basis of number of valence electrons.

Rb > Mg > Al

Reason : The reactivity is determined by the number of electrons present in the outermost shell that means the element which have 1 valence electron will be more reactive because they can easily lose electrons.

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1. Take the reaction: NH3 + O2 + NO + H2O. In an experiment, 3.25g of NH3 are allowed
Rudiy27

Answer:

5.74g of NO

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

4NH3 + 5O2 —> 4NO + 6H2O

Step 2:

Determination of the masses of NH3 and O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar Mass of NO = 14 + 16 = 30g/mol

Mass of NO from the balanced equation = 4 x 30 = 120g

From the balanced equation above,

68g of NH3 reacted with 160g of O2 to produce 120g of NO.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting because it will be used to calculate the maximum yield of the reaction. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Therefore, 3.25g of NH3 will react with = (3.25 x 160)/68 = 7.65g of O2.

From the simple illustration above, we can see that lesser mass of O2 is needed to react with 3.25g of NH3. Therefore, NH3 is the limiting reactant while O2 is the excess reactant.

Step 4:

Determination of the mass of NO produced from the reaction.

In this case the limiting reactant will be used because all of it were used in the reaction.

The limiting reactant is NH3.

From the balanced equation above,

68g of NH3 reacted to produce 120g of NO.

Therefore, 3.25g of NH3 will react to produce = (3.25 x 120)/68 = 5.74g of NO.

From the calculations made above, 5.74g of NO is produced.

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Not Significant:

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- Trailing zeroes with no decimal point: example- 2300= 2 sig figs

Significant:

- middle zeroes: example- 4003= 4 sig figs

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