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3 years ago

8

Which best describes an oxidizing agent?

2 answers:

stellarik [79]3 years ago

6 0

Answer:B) a reactant that undergoes reduction

Explanation:

Oxidation reaction is defined as the reaction in which a substance looses electrons. The oxidation state of the substance increases during oxidation.

$M\rightarrow M^++e^-$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced during reduction.

$N^++e^-\rightarrow N$

Overall reaction: $M+N^+\rightarrow M^++N$

The substance M which itself gets oxidized, reduces other and is called as reducing agent. The substance N which itself gets reduced, oxidizes other and is called as oxidizing agent.

victus00 [196]3 years ago

6 0

Answer:

b

Explanation:

cause it is

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Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/ato

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Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by $N_v$ .

It is depends on

total no. of atomic number(N)
energy required for vacancy
Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
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$N_v=Ne^{-\frac{Q_v}{kT} }$

To find equilibrium number of of vacancies we have find N.

$N=\frac{N_A\ \rho}{A_{cu}}$

Here ρ= 8.45 g/cm³ =8.45 ×10⁶m³

$N_A$ = Avogadro Number = 6.023×10²³

$A_{Cu}$ = 63.5 g/mole

$N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}$

$=8.01\times 10^{28$ g/mole

Here $Q_v$ =0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

$N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})$

=2.34×10²⁴ vacancies/ mole

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