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saveliy_v [14]
4 years ago
10

Fireworks, KNO3 Give only the names of the elements alphabetically, separating them with commas.

Chemistry
2 answers:
Alborosie4 years ago
4 0
Answer : The  question is to write the elements alphabetically and seperating them with commas.

So in this case for your question of Fireworks , KNO_{3},

The answer will be K, N, O.

You have to just differentiate the elements and write them according to the alphabetical order.
Ksenya-84 [330]4 years ago
4 0

<u>Answer:</u> The elements present are potassium, nitrogen and oxygen.

<u>Explanation:</u>

Elements are defined as the substances in which same type of atoms are present. <u>For Example:</u> Sodium, Chlorine etc.

The chemical formula of the compound used in fireworks is KNO_3

The chemical name for the compound is potassium nitrate.

This compound is formed by the combination of three elements. In 1 mole of potassium nitrate, 1 mole of potassium atom, 1 mole of nitrogen atom and 3 moles of oxygen atoms are present.

Hence, the elements present are potassium, nitrogen and oxygen.

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Answer:

- 622.5kJ

Explanation:

1)      2NH3 + 3N2O ----> 4N2 + 3H2O    ΔH° 1= - 1010kJ

-  3 (N2O  + 3H2     --->  N2H4 + H2O    ΔH° 2 = - 317 kJ)

2)      2NH3 + 3N2O ----> 4N2 + 3H2O        ΔH° 1= - 1010kJ

<u>      - 3N2O   - 9 H2     ----> - 3N2H4 - 3H2O      - 3ΔH° 2 = 3*317 kJ</u>

2NH3  + 3N2H4 --->4N2+9H2,     ΔH° 1 -  3*ΔH° 2

2NH3  + 3N2H4  --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

3) -(2NH3 +1/2O2 ---> N2H4 + H2O,     ΔH°3 = -143 kJ)

   -2NH3 - 1/2O2 ---> - N2H4 - H2O,    - ΔH°3 = 143 kJ

 2NH3  + 3N2H4 --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

<u>  -2NH3 - 1/2O2              ---> - N2H4 - H2O,             - ΔH°3 = 143 kJ</u>

 4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

4)

H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ

9*(H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ)

9H2+ 9/2 O2 ---> 9H2O,  9*ΔH° 4 = 9*(- 286) kJ

4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

<u>9H2+ 9/2 O2    ---> 9H2O,                             9*ΔH° 4 = 9*(- 286) kJ</u>

4N2H4 +4O2 --->4N2+8H2O,         ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

5)

1/4*(4N2H4 +4O2 --->4N2+8H2O,    ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

N2H4 +O2 --->N2+2H2O,        1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)

6)

N2H4 +O2 --->N2+2,        

1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=

=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ

   

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Explanation:

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7 0
3 years ago
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Answer: 2.12 L of NH3

Explanation:

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3 years ago
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