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2 years ago

When does the number of electrons be greater than the number of protons?And give examples.

Chemistry

1 answer:

vova2212 [387]2 years ago

4 0

Answer:

This accepting atom will then become an anion. The anion has more electrons than protons. This means that some of the electrons will not be canceled out, giving the atom an overall negative charge.

If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.

Explanation:

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Which statement correctly describes the relationship between reactant and yield? The actual yield is calculated from the amount

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2 years ago

Which ionic equation describes a redox reaction? A. Ag(+) + Cl- = AgCl B. 2H(+) + CO3(2-) = CO2 + H2O C. H(+) + OH(-) = H2O D. Z

Gennadij [26K]

Answer: The correct option is D. $Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously. It is known as the reaction in which the exchange of electrons takes place.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

From the given ionic reactions:

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

On the reactant side:

Oxidation number of Zn = 0

Oxidation number of Cu = +2

On the product side:

Oxidation number of Cu = 0

Oxidation number of Zn = +2

As the oxidation number of Zn is increasing from 0 to +2. Thus, it is getting oxidized. Similarly, the oxidation number of Cu is decreasing from +2 to 0. Thus, it is getting reduced. Therefore, forming a redox couple

Hence, the correct option is D. $Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

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2 years ago