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The CaCO3 produced if 47.5 moles of NH3 produced is calculated as follows
CaCN2 +3H2O = CaCO3 + 2NH3
by use of mole ratio between CaCO3 to NH3 which is 1:2 the moles of CaCO3 is therefore = 47.5 /2= 23.75 moles
mass of CaCO3 is therefore = moles x molar mass
= 23.75 moles x 100g/mol= 2375 grams which is approximate 2380 grams(answer 6)
The atomic radius decreases
Answer: I would say B.
Explanation:
Active transport requires energy, while passive does not.
