Question

The mole is a counting number that allows scientists to describe how individual molecules and atoms react. If one mole of atoms or molecules is equal to 6.022 x 10^32 atoms or molecules, how many molecules are in 23.45 g sample of copper (II) hydroxide, Cu(OH)2? Express your answer to the correct number of significant figures. (MM of Cu(OH)2 is 97.562g/mol. Be sure to show all steps completed to arrive at the answer.

Answer 1

Answer:

$\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}} }$

Explanation:

You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.

1. Moles of Cu(OH)₂$\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}$

2. Molecules of Cu(OH)₂$\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are \large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}} }$

Answer 2

The number of molecules of copper (II) hydroxide in 23.45 g sample has been $\rm \bold {1.45\;\times\;10^2^3}$.

According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been $\rm 6.023\;\times\;10^2^3$.

The moles of a compound has been given as:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

The moles in 23.45 g copper (II) hydroxide has been:

$\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol$

The moles of copper (II) hydroxide has been 0.24 mol.

The number of molecules in 0.24 mol sample has been driven by:

$\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules$

The number of molecules of copper (II) hydroxide in 23.45 g sample has been $\rm \bold {1.45\;\times\;10^2^3}$.

For more information about molecules in a mole of sample, refer to the link:

https://brainly.com/question/24577356