This is a disgusting mole problem and of which I have problem solving it in a clean manner.
Question 10:
Mercury has density of 13.6 g/mL. What is the volume of 1 atom of Mercury?
In order to get the point solve this using neat and tidy steps like moles cancel out with moles etc.. type of equation so I can learn. Thankyou.
2007-12-29 07:06:47 · 5 answers · asked by Ge1st 1 in Science & Mathematics ➔ Chemistry
Atomic weight: Hg=200.6
1mLHg/13.6gHg x 200.6gHg/1molHg x 1molHg/ 6.02x10^23atomsHg = 2.45 x 10^-23 mL/atom
2007-12-29 07:14:43 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
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2016-05-17 13:11:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
1 mL weights 13.6 grams
Moles = 13.6 g / 200.59 g/mol =0.0678 moles
Atoms = 0.0678 x 6.02 x 10^23 = 4.08 x 10^22
1 mL => 4.08 x 10^22 atoms
Volume of 1 atom = 1 / 4.08 x 10^22 = 2.45 x 10^-23 mL
2007-12-29 07:15:09 · answer #3 · answered by Dr.A 7 · 0⤊ 0⤋
200.59 g/mol (Atomic weight of Mercury)
13.6 g/mL (mass density)
6.02214 x 10^23 atoms/mol (Avogadro's number)
We want mL/atom
200.59 / 13.6 = 14.749 ml/mol
14.749 / (6.02214 x 10^23) = 2.4491 × 10-23 mL/atom
2007-12-29 07:25:20 · answer #4 · answered by Charlie149 6 · 0⤊ 0⤋
mass of one atom Hg is
13.6/6.022 x 10^23 = 2.26 x 10^-23g
This mass is divided by 1 ml
2.26 x 10^-23 / 1 = 2.26 x 10^-23 ml.
or
0.000,000,000,000,000,000,000,026 ml
2007-12-29 07:22:12 · answer #5 · answered by lenpol7 7 · 0⤊ 1⤋