Question

Q 12. Calculate the mass of 1000 molecules of calcium carbonate.
Q. 13. Calculate the number of moles present in a 8.5 g of Ammonia..​

Answer 1

Answer:

Ans12. mmu of CaCO3=100gms

6.022*10²³ molecules of CaCO3=100gmsof CaCO3

1 molecule of CaCO3=100gms/6.022*10²³

1000 molecules of CaCO3=100gms/6.022*10²³*1000

=1.66057788^-19

Ans13. mmu of NH3=17gms

17 gms of NH3= 1 mole of NH3

1gms of NH3= 1mole/17

8.5 gms of NH3= 1 mole/17*8.5

=1/2 mole of NH3

Explanation:

NOTE: In question 12 the value of answer is very small because the number of molecules are less.

Answer 2

Answer:

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