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3 years ago

If an atom of Bromine (Br) becomes an ion, which of the following does it most likely form?

Chemistry

1 answer:

lawyer [7]3 years ago

5 0

Answer:

An atom of Bromine (Br) forms an ion and becomes Br⁻

Explanation:

Atoms of elements gain or lose electron(s) to attain a stable configuration and form ions.
When an atom gains electron(s) it forms a negatively charged ion called an anion.
For example, Bromine is a halogen and its atom requires to gain one electron to attain stability and form a bromine ion (Br⁻).
When an atom loses electron(s) it forms a positively charged ion called a cation.
For instance, atoms of calcium(Ca) requires to lose two electrons to attain stability and form calcium ion (Ca²⁺).

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A dilute solution of bromine in carbon tetrachloride behaves as an ideal-dilute solution. The vapour pressure of pure CCl4 is 33

ANEK [815]

Explanation:

The given data is as follows.

Vapour pressure of pure $CCl_{4}$ = 33.85 Torr

Temperature = 298 K

Mole fraction of $Br_{2}$ = 122.36 torr

Therefore, calculate the vapor pressure of $Br_{2}$ as follows.

Vapour pressure of $Br_{2}$ = mole fraction of $Br_{2}$ x K of $Br_{2}$

= 0.050 x 122.36 Torr

= 6.118 Torr

So, vapor pressure of $Br_{2}$ is 6.118 Torr .

Now, calculate the vapor pressure of carbon tetrachloride as follows.

Vapour pressure of $CCl_{4}$ = mole fraction of $CCl_{4}$ x Pressure of $CCl_{4}$

= (1 - 0.050) × 33.85 Torr

= 32.1575 Torr

So, vapor pressure of $CCl_{4}$ is 32.1575 Torr .

Hence, the total pressure will be as follows.

= 6.118 Torr + 32.1575 Torr

= 38.2755 Torr

Therefore, composition of $CCl_{4}$ = $\frac{32.1575 Torr}{38.2755 Torr}$

= 0.8405

Composition of $CCl_{4}$ is 0.8405 .

And, composition of $Br_{2}$ = $\frac{6.118 Torr}{38.2755 Torr}$

= 0.1598

Composition of $Br_{2}$ is 0.1598 .

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