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Karolina [17]
4 years ago
6

Which of the following is chemically inert ( unreactive)?

Chemistry
1 answer:
IrinaK [193]4 years ago
4 0

(C) neon (atomic number 10)

Explanation:

The atom that is chemically inert is Neon with an atomic number of 10.

An atom is chemically inert if it does not willingly take part in chemical reaction.

Atoms takes part in chemical reactions in order to attain a stable configuration as seen in the noble gases.

In the noble gases every enable level is completely filled with the appropriate number of electrons.

  • The noble gases are He, Ne, Ar, Kr, Xe
  • These elements do not react with others.
  • From the given option, Neon is a noble gas and therefore inert.

learn more:

Noble gas brainly.com/question/1781595

#learnwithBrainly

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Explanation:

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Divergent boundaries – where two plates are moving apart. ...

Transform boundaries – where plates slide passed each other.

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What is Aqua Auras mineral family/class?
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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

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