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lubasha [3.4K]
2 years ago
15

Which of the following shows an accurate combustion reaction?

Chemistry
1 answer:
Jet001 [13]2 years ago
5 0

Answer:

  • <em>Cu + O₂  → CuO₂</em>

Explanation:

A <em>combustion reaction</em> is the reaction with oxygen along with the release of energy in form of heat or light.

Organic compounds (like CH₄) undergo combustion forming water and CO₂.

The combustion reaction of CH₄ is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

Hence, the first equation from the choices is not showing the combustion reaction of CH₄.

Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.

The reaction of copper and oxygen (second choice) is a combustion reaction:

  • <em>Cu + O₂ → CuO₂</em>

The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.

On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.

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Answer:

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Explanation:

Exothermic is a process in which heat is released during the process. Endothermic reactions absorbs heat from surrounding during a chemical process. The dissolution of salt into water is an exothermic reaction. During this process heat is release and water molecules are broken down which are surrounded by salt ions.

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5 0
2 years ago
Read 2 more answers
What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)
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Explanation:

The given data is as follows.

            Energy of radiation absorbed by the electron in hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

                                    E = h \nu

               1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

               \nu = 0.163 \times 10^{17} s^{-1}

or,                \nu = 1.63 \times 10^{16} s^{-1}    

It is known that,        \nu = \frac{c}{\lambda}

                1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}                  

                   \lambda = 1.84 \times 10^{-8} m

And, according to De-Broglie equation \lambda = \frac{h}{p}

as,        p = m \times \nu

So,          \lambda = \frac{h}{m \times \nu}

            m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}          

                             = 3.6 \times 10^{-26} J/m

Now, on squaring both the sides we get the following.

           (m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}    

                              = 12.96 \times 10^{-52}  

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where,   m = mass of electron

So,           m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

                             = \frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}

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Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is 7.1 \times 10^{-22} J.

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