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Dmitry [639]
3 years ago
15

46. True or false. Chemical reactions do not involve changes in the chemical bonds that join

Chemistry
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

Chemical reactions do not involve changes in the chemical bonds that join

atoms in compounds :

<u>False</u>

Explanation:

Chemical reaction are the reaction in which old bonds break and new  bonds are formed . The formation of new bond result in formation of new compounds . This happen because new bond are result of linking different atoms by the bond.

For example : Water formation from Oxygen and Hydrogen is a chemical process :

2H_{2}+O_{2}\rightarrow 2H_{2}O

Original(old) bonds are :

H-H bond in H2 and O-O bonds in O2

In H2 = Hydrogen is joined to Hydrogen

IN O2 = Oxygen is joined to oxygen

New Bonds =

O-H bonds in water (H2O)

Oxygen is joined to hydrogen = New Bond formation

Hence,

<u>Chemical reactions do involve changes in the chemical bonds that join </u>

<u>atoms in compounds</u>

<u />

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The gravitational potential energy it had from being above the ground is converted to kinetic energy as the rock falls. As kinetic energy increases, the velocity of the rock will also increase. However, if one considers the air, the rock will lose energy as it falls due to air resistance. If the cliff is high enough, the acceleration due to gravity and the air resistance due to the velocity will be exactly equal and there will be no change in kinetic energy.

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7 0
2 years ago
Enter your answer in the provided box. Oxidation of gaseous ClF by F2 yields liquid ClF3, an important fluorinating agent. Use t
MakcuM [25]

Answer : The value of \Delta H_{rxn}  for the reaction is, -135.2 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.

The formation of ClF_3 will be,

ClF(g)+F_2(g)\rightarrow ClF_3(l)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) 2ClF(g)+O_2(g)\rightarrow Cl_2O(g)+OF_2(g)     \Delta H_1=167.5kJ

(2) 2F_2(g)+O_2(g)\rightarrow 2OF_2(g)    \Delta H_2=-43.5kJ

(3) 2ClF_3(l)+2O_2(g)\rightarrow Cl_2O(g)+3OF_2(g)    \Delta H_3=394.4kJ

We are dividing the reaction 1, 2 and 3 and reversing reaction 3 and then adding all the equations, we get :

(1) ClF(g)+\frac{1}{2}O_2(g)\rightarrow \frac{1}{2}Cl_2O(g)+\frac{1}{2}OF_2(g)     \Delta H_1=\frac{167.5kJ}{2}=83.75kJ

(2) F_2(g)+\frac{1}{2}O_2(g)\rightarrow OF_2(g)    \Delta H_2=\frac{-43.5kJ}{2}=-21.75kJ

(3) \frac{1}{2}Cl_2O(g)+\frac{3}{2}OF_2(g)\rightarrow ClF_3(l)+O_2(g)    \Delta H_3=\frac{-394.4kJ}{2}=-197.2kJ

The expression for enthalpy of formation of ClF_3 will be,

\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(83.75kJ)+(-21.75kJ)+(-197.2kJ)

\Delta H_{rxn}=-135.2kJ

Therefore, the value of \Delta H_{rxn}  for the reaction is, -135.2 kJ

5 0
3 years ago
A catalyst lowers the activation energy for both the forward and the reverse reactions in an equilibrium system, so it has no ef
rosijanka [135]

Answer:

True.

Explanation:

  • Catalyst increases the rate of the reaction without affecting the equilibrium position.
  • Catalyst increases the rate via lowering the activation energy of the reaction.
  • This can occur via passing the reaction in alternative pathway (changing the mechanism).
  • The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
  • in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
  • with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.
  • Kindly, see the attached image.

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Hence; 0.0128 moles × 6.022 ×10^23 molecules
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