The radius of gold is 144 picometers, so....,
144 picometers (103/1012)
= 1.44 X 10 -7 mm
------------------------so,
4.0 mm/1.44 X 10 -7 mm
= 2.8 X 107 * 2 ( because you need to double radius to get diameter end to end )
= 5.6 X 107 gold atoms lined up
======================
The answer is 9,17233.1023 atoms.
One Angstrom is defined as 10-10 centimeters. Therefore, the number of any units with a diameter of 2.2 Angstroms that would be required to span 1.0 cm is 1.0/10 -10 or 4.5 X 109 calcium atoms., to the justifiable number of significant digits.
6.14x1019 atoms Au
This depends on the mass of the gold sample.
6.02*10(to the 23rd)
Gold is the element gold no matter how many atoms of it you have.
If there is one sulphate molcule, there will be 2 gold atoms.
7875648/4754878748485456487875456152326599789796561245849849796566165456484545423323485465454580084548545215454845151458487545234154845456645465456454534noob56587988565954954656549494664seriously?4156646546545646546545645646546546543212378945452465987
The answer is 9,17233.1023 atoms.
One Angstrom is defined as 10-10 centimeters. Therefore, the number of any units with a diameter of 2.2 Angstroms that would be required to span 1.0 cm is 1.0/10 -10 or 4.5 X 109 calcium atoms., to the justifiable number of significant digits.
6.14x1019 atoms Au
This depends on the mass of the gold sample.
This depends on the mass of the gold sample.
6.02*10(to the 23rd)
4.2 X 10^24 atoms of gold / 6.022 X 10^23 atoms per mol) = 7.0 moles of gold
Yes. One mole (6.02 x 10 to the 23 atoms) would weigh 197 grams.
250 atoms