if the diameter of chromium atom is about 2.4A then how many chromium atoms would have to be lined up to span of 1 cm?
An angstrom = 1 X 10^-10 meters 1.0 mm = 0.001 meters so, 1.9 * 1 X 10^-10 = 1.9 X 10^-10 meter per chromium so 0.001 meters ( 1 meter/1.9 X 10^-10) = 5263158 atoms of chromium ------------------------------------------ ( 5.3 X 10^6 atoms )
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 atomic radii of an Au atom is 174 pm. A pico meter is 10^-12 of a meter. Use this to easily convert to mm. Then multiply this value by two to find the diameter of one Au atom in mm. Finally, divide 1.5 by this answer to find the total Au atoms to span. You should get approximately 4.45 x 10^7 Au atoms.
To find the number of chromium atoms in 147.4 kg of chromium, we first need to convert the mass to moles using the molar mass of chromium (51.996 g/mol). Then, we can use Avogadro's number (6.022 x 10^23 atoms/mol) to find the number of atoms. The calculation would be (147.4 kg / 51.996 g/mol) * 6.022 x 10^23 atoms/mol.
The radius of krypton is 110 picometers, which is 220 picometers diameter, or the length we want of the Kr atom. So, convert 3.5 millimeters (10^9/10^3) = 3500000 picometers/220 picometers = 15,909 atoms of Krypton
An angstrom = 1 X 10^-10 meters 1.0 mm = 0.001 meters so, 1.9 * 1 X 10^-10 = 1.9 X 10^-10 meter per chromium so 0.001 meters ( 1 meter/1.9 X 10^-10) = 5263158 atoms of chromium ------------------------------------------ ( 5.3 X 10^6 atoms )
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======================
1,000,000,000 atoms lined up side by side will equal 1 meter * * * * * I disagree. The diameter of a copper atom is approx 250 picometres = 2.5*10-10 metres. You would therefore require 4 billion atoms.
The atomic radii of an Au atom is 174 pm. A pico meter is 10^-12 of a meter. Use this to easily convert to mm. Then multiply this value by two to find the diameter of one Au atom in mm. Finally, divide 1.5 by this answer to find the total Au atoms to span. You should get approximately 4.45 x 10^7 Au atoms.
1,000,000,000 atoms lined up side by side will equal 1 meter * * * * * I disagree. The diameter of a copper atom is approx 250 picometres = 2.5*10-10 metres. You would therefore require 4 billion atoms.
To find the number of chromium atoms in 147.4 kg of chromium, we first need to convert the mass to moles using the molar mass of chromium (51.996 g/mol). Then, we can use Avogadro's number (6.022 x 10^23 atoms/mol) to find the number of atoms. The calculation would be (147.4 kg / 51.996 g/mol) * 6.022 x 10^23 atoms/mol.
The radius of krypton is 110 picometers, which is 220 picometers diameter, or the length we want of the Kr atom. So, convert 3.5 millimeters (10^9/10^3) = 3500000 picometers/220 picometers = 15,909 atoms of Krypton
Chromium nitrate is an ionic compound. It is composed of chromium cations (Cr3+) and nitrate anions (NO3-), which are held together by ionic bonds due to the transfer of electrons from chromium to nitrate.
It would take about 254 billion Carbon atoms lined up in a single row to span one inch.
It would take about 109 Earths lined up side by side to span the diameter of the sun. The sun is incredibly large, with a diameter of about 1.4 million kilometers, compared to Earth's diameter of just over 12,000 kilometers.
100,000,000 copper atoms4 or 5 billionAccording to the link, the diameter of a copper atom is 2.28e-10 m, or 2.28x10-10m. The answer to your question is 1/(2.28x10-10). this can be rewritten as follows:(1/2.28)(1/10-10) = (approx. 0.439)(1010) = 4.39x109.You could say "over 4 billion".
About half of a centimeter, more or less depending on what elements the atoms were.