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dsp73
3 years ago
15

How many grmas of CaCO3 are present in a sample if there are 4. 52 x 10^24 atoms of carbon in that sample?

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
3 0

No of atoms in 1 mol of CaCO3=6.023×10^23

\\ \sf\bull\longmapsto No\:of\:moles=\dfrac{4.52\times 10^24}{6.023\times 10^23}

\\ \sf\bull\longmapsto No\:of\:moles=0.7mol

Molar mass of CaCO3

\\ \sf\bull\longmapsto 40u+12u+3(16u)

\\ \sf\bull\longmapsto 52u+48u

\\ \sf\bull\longmapsto 100g/mol

Now

\\ \sf\bull\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}

\\ \sf\bull\longmapsto Given\:mass=No\:of\:moles\times Molar\:Mass

\\ \sf\bull\longmapsto Given\:Mass=0.7(100)

\\ \sf\bull\longmapsto Given\:Mass=70g

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Rate\ \alpha [A]^{a}[B]^{b}

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STEP": First, find out the power "a" and "b"

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According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

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<u>r' = 1/4 r</u>

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