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\[1.20 \times 10^{22} atoms \times \frac{1mol}{6.02\times 10^{23}atoms}\times \frac{200g}{1mol}=?\]
4?
yes
.Sam. has used a periodic table to find \[A_r[\text{Hg]}=200.59\left[\frac{\text{g}}{\text{mol}}\right]\]
''Please explain how you came to your answer'' First find the no. of mole of Hg there. No. of mole Hg = no. of atoms / Avogadro constant = 1.20 × 10^22 / 6.02x10^23 = 0.0199 mole Mass of Hg = no. of mole x atomic mass of Hg = 0.0199 x 200.59 = 4g
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