Discover new perspectives and gain insights with IDNLearn.com. Get prompt and accurate answers to your questions from our experts who are always ready to help.

Calculate the number of cations and anions in each of the following compounds. Enter your answers in scientific notation. (a) 6.42 g of KBr:

Sagot :

Answer: The number of cations are [tex]3.24 \times 10^{22}[/tex] and number of anions are  [tex]3.24 \times 10^{22}[/tex] in 6.42 g of KBr.

Explanation:

The molar mass of KBr is (39.10 + 79.90) g/mol = 119.00 g/mol

Now, the dissociation equation for KBr is as follows.

[tex]KBr \rightarrow K^{+} + Br^{-}[/tex]

This means that 1 mole of KBr is forming 1 mole of [tex]K^{+}[/tex] (cation) and 1 mole of [tex]Br^{-}[/tex] (anion).

According to mole concept, 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms. Hence, number of cations present in 6.42 g KBr is calculated as follows.

[tex]No. of cations = Moles \times 6.022 \times 10^{23}\\= \frac{mass}{molar mass} \times 6.022 \times 10^{23}\\= \frac{6.42 g}{119.00 g/mol} \times 6.022 \times 10^{23}\\= 3.24 \times 10^{22}[/tex]

As according to the equation, there are equal number of moles of both cation and anions.

This means that the number of anions are also [tex]3.24 \times 10^{22}[/tex].

Thus, we can conclude that the number of cations are [tex]3.24 \times 10^{22}[/tex] and number of anions are  [tex]3.24 \times 10^{22}[/tex] in 6.42 g of KBr.

The number of cations and anions present in potassium bromide is 3.24 × 10²².

How we calculate atoms from moles?

In one mole of any substance 6.022 × 10²³ atoms of that substance is present and this is known as Avogadro's number.

KBr is a strong electrolyte means it fully dissociates into their constitute ions. So, the number of moles of produced ions is equal to the moles of KBr and dissociation is represented as:

KBr → K⁺ + Br⁻

From this it is clear that 1 mole of cation and 1 mole of anion is produced from 1 mole of KBr. Moles of KBr will be calculated as:

n = W/M, where

W = given mass = 6.42g

M = molar mass = 119

n = 6.42 / 119 = 0.053 moles

No. of cations and anions present in 0.053 moles = 0.053 × 6.022 × 10²³ = 3.24 × 10²².

Hence, 3.24 × 10²² is the no. of cations and anions.

To know more about Avogadro's number,  visit the below link:
https://brainly.com/question/10614569