hfjdhdhsus hfjdhdhsus
  • 16-03-2021
  • Chemistry
contestada

Find the moles of CaCO3 in 23.4 g of calcium carbonate

Respuesta :

aniyabacchus aniyabacchus
  • 16-03-2021

Answer:

0.0326 moles

explanation

Answer Link
bennysar bennysar
  • 16-03-2021

Answer:

0.234 mol

Explanation:

m(CaCO3) = 23.4g

n(CaCO3) = ?

M(CaCO3) = 100.09 g/mol

n = m/M

n(CaCO3) = 23.4 / 100.09

                 = 0.234 mol (3 s.f.)

Answer Link

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