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Triss [41]
3 years ago
6

How many molecules are found in 90 grams of Al(OH)3?

Chemistry
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

6.95 x 10²³ molecules/particles

Explanation:

First we need to find the total Empirical Mass. We can find this by adding each element's mass together.

Al = 26.982,

O = 15.999

H = 1.008

26.982 + 3(15.999) + 3(1.008) = 78.003.

Now we divide by the mass given (90 grams).

90/78.003 = 1.153801777.

We then take that number and multiply it by avogadro's number (6.022 x 10²³)

1.153801777 x avogadro's number = 6.95 x 10²³

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Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith
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The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
  • Atomic mass of lithium = 6.94 u
  • Abundance of A = A%
  • Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

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6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

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Abundance of B = (100 - 8)%

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  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

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brainly.com/question/24311846

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