In the presence of CN-, Fe3 forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3 and Fe(CN)63- are 8.5 x 10-40
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Answer:
A) Sample B has more calcium carbonate molecules
Explanation:
M = Molar mass of calcium carbonate = 100.0869 g/mol
= Avogadro's number =
For the 4.12 g sample
Moles of a substance is given by
Number of molecules is given by
For the 19.37 g sample
Number of molecules is given by
So, sample B has more calcium carbonate molecules.
The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.