plz answer this question asap plz plz fast
Q. no. 9 & 10
if you know the answer then only reply or I'll report your answer
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please mark it as brainliest
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answer is 5u of 5 atoms.
Explanation:
Molecular weight of CaCO3 is 100. So, number of moles present in 20g of CaCO3 are 20/100=0.2. Each molecule of CaCO3 has 5 atoms. So, total no. of atoms present are 0.2×5.Na=6.023×10^23.
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