Calcute the number of alluminum irons present in 0.0051 g of aluminum

1.	Calcute the number of alluminum irons present in 0.0051 g of aluminum oxide . ( hint: the mass of an ion is the same as that of an atom of the element. Atomic number of aluminum 27) 5 mark
Answer» Answer: Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (HINT: The mass of an ion is the same as that of an atom of the same element. Atomic mass of AL = 27 u) . Solution: Mole of aluminium oxide (Al2O3) = 2 x 27 + 3 x 16 = 102 g i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3 Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / 102 x 0.051 molecules = 3.011 x 1020 molecules of Al2O3 The number of aluminium ions (AL3+) present in one molecule of aluminium oxide is 2. Explanation: HOPE THIS HELPS YOU BETTER PLZ MARK ME AS BRAINLIEST GIVE THANKS =TAKE THANKS

Calcute the number of alluminum irons present in 0.0051 g of aluminum oxide . ( hint: the mass of an ion is the same as that of an atom of the element. Atomic number of aluminum 27) 5 mark

Answer»

Answer:

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

(HINT: The mass of an ion is the same as that of an atom of the same element. Atomic mass of AL = 27 u) .

Solution:

Mole of aluminium oxide (Al2O3) = 2 x 27 + 3 x 16

= 102 g

i.e., 102 g of Al2O3= 6.022 x 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / 102 x 0.051 molecules

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (AL3+) present in one molecule of aluminium oxide is 2.

Explanation:

HOPE THIS HELPS YOU BETTER

PLZ MARK ME AS BRAINLIEST

GIVE THANKS =TAKE THANKS

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