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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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Explanation:

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Balance the following equation:<br> K+ H2O + H2 + KOH
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Answer:

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Explanation:

Find how many atoms you have on both sides then add 2 to both sides.

Reactant:       Products:

K: 1+1=2                K: 1+1=2

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6 0
2 years ago
Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

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Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

5 0
2 years ago
Consider the reaction at 500 ° C 500°C . N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) K c = 0.061 N2(g)+3H2(g)↽−−⇀2NH3(g)Kc=0.06
Sav [38]

Answer:

Q = 0.061 = Kc

Explanation:

Step 1: Data given

Temperature = 500 °C

Kc=0.061

1.14 mol/L  N2

5.52 mol/L H2

3.42 mol/L NH3

Step 2: Calculate Q

Q=[products]/[reactants]=[NH3]²/ [N2][H2]³

If Qc=Kc then the reaction is at equilibrium.  

If Qc<Kc then the reaction will shift right to reach equilibrium.

If Qc>Kc then the reaction will shift left to reach equilibrium.  

Q = (3.42)² / (1.14 * 5.52³)

Q = 11.6964/191.744

Q = 0.061

Q = Kc the reaction is at equilibrium.  

4 0
3 years ago
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