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dybincka [34]
3 years ago
14

If we read the reaction as X + Y → Z + Q it is A. endothermic B. exothermic

Chemistry
2 answers:
Korvikt [17]3 years ago
4 0
As you can see in the picture we have +ΔH so that means for this reaction we need to GET heat. so the answer is A. endothermic :))
i hope this is helpful
have a nice day 
kupik [55]3 years ago
3 0

Answer: The given reaction is a type of endothermic reaction.

Explanation:

There are 2 types of reaction classified on the basis of heat released or absorbed:

1. Exothermic reactions: In these reactions, the heat is released because the energy of products is less than the energy of the reactants. \Delta H for these reactions is negative.

2. Endothermic reactions: In these reactions, heat is absorbed by the reactants because the energy of the products is more than the energy of the reactants. \Delta H for these reactions if positive.

In the given question, the energy of the products is greater than the energy of the reactants. Hence, this is considered as a type of endothermic reaction.

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Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle
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Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71\times 10^{-15}J . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

= De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of beta particle = 9.1094\times 10^{-31} kg

E_k = kinetic energy of the particle = 4.71\times 10^{-15}J

Putting values in above equation, we get:

\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}

\lambda = 6.762\times 10^{-12} m

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

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