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Vlad [161]
2 years ago
13

If a collision between molecules is very gentle the molecules are

Chemistry
1 answer:
Nadya [2.5K]2 years ago
6 0
If  collision  between   molecules  is  very gentle  the   molecules  are  more   likely  to   bound  without  reaction

     c
ollision   is  process  in  which  molecules  come  together   or  collide   with  one  another.Molecules  must  collide  with  sufficient   energy  so  that   chemical  bond can break.  In  addition  for  collision  to  occur  there  must  have  favorable  orientation.
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2 years ago
Given the following reactions Fe2O3 (s) + 3CO (s) → 2Fe (s) + 3CO2 (g) ΔH = -28.0 kJ 3Fe (s) + 4CO2(s) → 4CO (g) + Fe3O4(s) ΔH =
Taya2010 [7]

Answer: The enthalpy of the reaction is -109 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Fe_2O_3(s)+3CO(s)\rightarrow 2Fe(s)+3CO_2(g)   \Delta H=-28.0kJ  \times 3    

3Fe_2O_3(s)+9CO(s)\rightarrow 6Fe(s)+9CO_2(g)   \Delta H=-84.0kJ     (1)

3Fe(s)+4CO_2(s)\rightarrow 4CO(g)+Fe_3O_4(s) \Delta H=+12.5kJ  \times 2  

6Fe(s)+8CO_2(s)\rightarrow 8CO(g)+2Fe_3O_4(s) \Delta H=+25.0kJ     (2)

The final reaction is:

Subtracting (2) from (1):

3Fe_2O_3(s)+CO(g)\rightarrow CO_2(g)+2Fe_3O_4(s) \Delta H=-84.0-(+25.0)=-109kJ

Thus the enthalpy of the reaction is -109 kJ

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2 years ago
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Hope this helps! Please correct me if I'm wrong :)

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balance the following reaction using LCM method by showing each steps Pb (N3)2 + Cr(MnO4)2  Cr2O3 + MnO2 + Pb3O4+ NO​
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here's the answer to your question

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