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stiv31 [10]
3 years ago
8

How does the electron configuration of elements within the same group compare? (A.) They all have their valence electrons in the

same energy level. ( B) They all have their valence electrons in the same type of subshell. (C) They all have their valence electrons in a p suborbital. ( D) They all have their valence electrons in the 3rd energy level.
Chemistry
1 answer:
lesya [120]3 years ago
5 0

Answer:

B.They all have their valence electrons in the same type of subshell

Explanation:

  • The electron configurations of elements in the same group (column) of the periodic table have them in the same type of subshell.
  • But the subshells may be of different shells. Thus , the energies of them need not be the same.
  • For example , The Alkalai Metals are found in the first column of the periodic table Group IA. This set of elements all have valence electrons in only the 's' orbital and because they are in the first column they all have s^{1} configuration. i.e,
  1. H 1s^1
  2. Li 1s^2 2s^1
  3. Na 1s^2 2s^2 2p^6 3s^1
  4. K 1s^2 2s^2 2p^6 3s^2 3p^64s^1
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When light waves pass through solids, the speed of the wave ___________.
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A sample of 23.2 g of nitrogen gas is reacted with
slavikrds [6]

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

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The answer is

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