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jasenka [17]
3 years ago
15

How many atoms are present in 0.10mol of ptcl2 (nh3)2?

Chemistry
2 answers:
Alexus [3.1K]3 years ago
7 0

Explanation:

According to the mole concept, 1 mole of an atom contains 6.022 \times 10^{23} atoms or molecules.

Therefore, we can calculate the number of atoms present in 0.10 mol of PtCl_{2}(NH_{3})_{2} as follows.

               0.10 mol \times 6.022 \times 10^{23} atoms

                   = 0.6022 \times 10^{23}

                   = 6.022 \times 10^{22}

Therefore, we can conclude that there are 6.022 \times 10^{22} atoms present in 0.10 mol of PtCl_{2}(NH_{3})_{2}.

serious [3.7K]3 years ago
3 0
The number of atoms  present in 0.10 mol of PtCl2(NH3) is calculated using Avogadro law  formula

that is

1 mole = 6.02 x10^23 atoms
what about 0.10 moles = ? atoms

= 6..02 x10 ^23 x 0.1 = 6.02 x10^22  atoms
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