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kap26 [50]
3 years ago
6

The picture on the left represents the reactants in the equation and the picture on the right represents the products. Does this

reaction follow the law of conservation of matter?

Chemistry
1 answer:
JulijaS [17]3 years ago
6 0
According to Law of conservation of matter," matter can neither be created nor destroyed but is conserved and remains constant over time'.

In above picture let suppose the Blue balls represent N₂ molecule and White balls represent H₂ molecules.

So, left picture represent reactants,

                                    2 N₂  +  6 H₂

And , right picture represent products,

                                     4 NH₃

So, there are 4 N atoms and 12 Hydrogen atoms in reactants and 4 N atom and 12 Hydrogen atoms in products. Means the mass of elements is conserved. The overall reactions is as follow,


                                    2 N₂  +  6 H₂   →   4 NH₃

Result:
          Yes! This reaction follow Law of conservation of Matter.
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the atomic number of barium is 56 what do you know about the subatomic particles in an atom of this element
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Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0
3 years ago
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