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Step2247 [10]
3 years ago
8

Which of the following is an exothermic reaction?

Chemistry
2 answers:
alekssr [168]3 years ago
7 0

Answer:

C.) condensing steam

Explanation:

Exothermic reactions are chemical reaction which energy is released to the environment in the form of heat . The overall energy of the reactant is greater than the energy of the product. The higher in energy of the reactant causes a release of heat to the environment. In exothermic reactions, enthalpy change is  usually negative and is accompanied by the loss of energy in the form of heat or light.

Condensation is the change of water vapor to liquid . The conversion is an exothermic reaction as energy is release to the environment in the form of heat.  The water vapor molecules has higher energy than liquid. The changes of water vapor to liquid causes a loss in energy of the water vapor. As the water vapor loss more energy, heat is been release to the surrounding.

All other reaction are endothermic in which heat is absorb from the environment except for condensation of steam.

PilotLPTM [1.2K]3 years ago
6 0

The event that is an example of an exothermic reaction is condensing steam. The answer is letter C. An exothermic reaction is one in which there is a release of heat during the reaction. Dissolving sugar in water, melting ice and dissolving ammonium nitrate in water are all examples of an endothermic reaction.

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Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸  J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} =   e ^{^{-\dfrac{(\Delta E)}{KT}}}

From the above equation:

Δ E = energy difference =  3.13 × 10⁻¹⁸  J

k = Boltzmann constant = 1.38 × 10⁻²³  J/K

\dfrac{N_2}{N_1} = 0.5

Thus:

0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}

-3.00 = -226811.5942 \times \dfrac{1}{T}

T =  \dfrac{-226811.5942}{-3.00 }

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