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)
Molar Mass of Al2O3 = 2 x Atomic mass of Al + 3 x Atomic mass of O
= 2 x 27 + 3 x 16
= 54 + 48 = 102 g
102 g of Al2O3 contain = 2 x 6.022 x 1023 aluminium ion
= 0.051 g of Al2O3 = 2 x 6.022 x 1023 / 102 x 0.051
= 6.022 x 1023 x 10-3 = 6.022 x 1020
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium observations are in agreement with the law of conservation of mass.
sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide
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(i) H2S molecule and
(ii) PO43- ion?
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Who discovered cells, and how?
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For increasing production, what is common in poultry, fisheries and bee-keeping?
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