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jonny [76]
3 years ago
15

All the substances listed below are fertilizers that contribute nitrogen to the soil.

Chemistry
1 answer:
kvv77 [185]3 years ago
4 0

Answer:

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

Explanation:

Percentage of element in compound :

=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100

(a) Urea, (NH_2)_2CO

Molar mass of urea = 60 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

N\%=\frac{2\times 14 g/mol}{60 g/mol}\times 100=46.67\%

(b) Ammonium nitrate, NH_4NO_3

Molar mass of ammonium nitrate = 80 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

N\%=\frac{2\times 14 g/mol}{80 g/mol}\times 100=35.00\%

(c) Nitric oxide, NO

Molar mass of nitric oxide  = 30 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

N\%=\frac{1\times 14 g/mol}{30 g/mol}\times 100=46.67\%

(d) Ammonia, NH_3

Molar mass of ammona = 17 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

N\%=\frac{1\times 14 g/mol}{17 g/mol}\times 100=82.35\%​

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

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A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513
igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is:  N₂O₄(g)  ⇆  2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

              2NO₂ (g)   ⇆   N₂O₄(g)      

Initially   2.20 mol              -

React          x                      x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq       (2.20-x) / 3.50L     (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) /  [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0  → Quadratic formula

a = 0.2933 ;  b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50  = 0.0971 M

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