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Answer :
Answer:
[tex]\boxed {\boxed {\sf About 7.1 \ mol \ H_2O}}[/tex]
Explanation:
To convert from molecules to moles, we must use Avogadro's Number: 6.022*10²³. This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are molecules of water.
[tex]\frac {6.022*10^{23} \ molecules \ H_2O} {1 \ mol \ H_2O}[/tex]
Multiply by the given number of molecules.
[tex]4.3 *10^{24} \ molecules \ H_2O *\frac {6.022*10^{23} \ molecules \ H_2O} {1 \ mol \ H_2O}[/tex]
Flip the fraction so the molecules of water cancel.
[tex]4.3 *10^{24} \ molecules \ H_2O *\frac {1 \ mol \ H_2O} {6.022*10^{23} \ molecules \ H_2O}[/tex]
[tex]4.3 *10^{24} *\frac {1 \ mol \ H_2O} {6.022*10^{23} }[/tex]
[tex]\frac {4.3 *10^{24}\ mol \ H_2O} {6.022*10^{23} }[/tex]
[tex]7.140484889 \ mol \ H_2O[/tex]
The original measurement of atoms has 2 significant figures ( 4 and 3), so our answer must have the same. For the moles we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.
[tex]7.1 \ mol \ H_2O[/tex]
There are about 7.1 moles of water.
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Rewritten by : Barada
There are [tex]7.14[/tex]moles in [tex]4.3 \times 10^{24}[/tex] molecules of [tex]H_2O[/tex] which is calculated using the Avogadro`s number .
To determine how many moles are in [tex]4.3 \times 10^{24}[/tex] molecules of [tex]H_2O[/tex], we must utilize Avogadro's formula., which is [tex]6.022 \times 10^{23}[/tex] molecules per mole. This number helps us convert between the number of molecules and moles.
- First, let's set up the conversion:
Number of moles = (Number of molecules) / (Avogadro's number)
- Now, we substitute the given values:
Number of moles = [tex]\frac{4.3 \times 10^{24} \, \text{molecules}}{6.022 \times 10^{23} \, \text{molecules/mol}}[/tex]
- Dividing the numbers, we get:
Number of moles = 7.14 moles
Therefore, [tex]4.3 \times 10^{24}[/tex] molecules of [tex]H_2O[/tex] equal approximately [tex]7.14[/tex]moles.
Additionally, knowing the relationship expressed by Avogadro's number is essential in various calculations within chemistry, from determining molecular masses to understanding chemical reactions and stoichiometry.