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How many formula units are in 98.3 grams of Al(OH)3?

Answer :

Answer:

78.003558

Explanation:

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Final answer:

To find the number of formula units in 98.3 grams of Al(OH)3, divide the given mass by the molar mass to find the number of moles, and then multiply by Avogadro's number to find the number of formula units.

Explanation:

To find the number of formula units in 98.3 grams of Al(OH)3, we need to use the concept of moles and Avogadro's number. First, we need to calculate the moles of Al(OH)3 using its molar mass. The molar mass of Al(OH)3 is calculated by adding the atomic masses of Al, O, and H, which are 26.98 g/mol, 16.00 g/mol, and 1.01 g/mol respectively. Therefore, the molar mass of Al(OH)3 is 78.00 g/mol. Next, we can calculate the number of moles of Al(OH)3 by dividing the given mass (98.3 g) by the molar mass (78.00 g/mol). This gives us approximately 1.26 moles of Al(OH)3.

Now, we can use Avogadro's number, which is 6.02 x 10^23 formula units per mole, to find the number of formula units in 1.26 moles of Al(OH)3. Multiplying 1.26 moles by Avogadro's number, we get approximately 7.59 x 10^23 formula units.

Therefore, there are approximately 7.59 x 10^23 formula units in 98.3 grams of Al(OH)3.