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Gekata [30.6K]
3 years ago
12

calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?

Chemistry
1 answer:
Assoli18 [71]3 years ago
3 0

The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.

<u>Explanation:</u>

An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.

So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as c a^{+} ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to F^{-} ions. This process of acceptance of electrons is termed as reduction.

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Explanation :

Density : The mass per unit volume of a substance is known as density.

Formula used:

\text{Density}=\frac{\text{Mass}}{\text{Volume}}

As we are given:

Density of mercury = 13.5 g/mL

Mass = 12.5 pounds

First we have to convert mass of sample from pound to gram.

Conversion used:

As, 1 pound = 453.6 g

So, 12.5 pounds = 453.6 × 12.5 g = 5670 g

Now we have to calculate the volume of liquid.

\text{Density}=\frac{\text{Mass}}{\text{Volume}}

Now putting all the given values in this formula, we get:

13.5g/mL=\frac{5670g}{\text{Volume}}

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