Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
<span>4NH</span>₃<span> + 6NO → 5N</span>₂<span> + 6H</span>₂<span>O
mol of NO = </span>
=
= 0.93 mol
Based on the balance equation mole ratio of NH₃ : NO is 4 : 6
= 2 : 3
If mol of NO = 0.93 mol
then mol of NH₃ =
= 0.62 mol
Mass of ammonia = mol × molar mass
= 0.62 mol × 17.03 g/mol
= 10.54 g
Therefore B is the best answer
Electron Configuration is the distribution of electrons of an atom or molecule in an atomic or molecular orbitals.
So Na, or Sodium has 11 electrons, so the number should add up to 11
Na: 1s² 2s² 3p⁶ 3s¹
Hope this helps!