The answer is 800,424.1026 atoms.
1 mole or 4 g or 0.004 kg of Helium has 6 x 1023 atoms. So, 431 kg of Helium have 65 x 1028 atoms
1 mole of helium (or 4 g or 0.004 kg) has 6 x 1023 atoms. So, 590 kg of helium will have 8.9 x 1028 atoms
The answer is 50,38349e+23 atoms.
12,4439 kg of gold contain 63,177 moles.
Here are the densities at STP: Air (ρair) = 1.292 kg/m3. Helium (ρHe) = 0.178 kg/m3 so you'd need to compress Helium about 7x to achieve the same density as air.
1 mole of helium (or 4 g or 0.004 kg) has 6 x 1023 atoms. So, 536 kg of helium will have 8.04 x 1028 atoms
The answer is 8,1547.10e25.
1 mole or 4 g or 0.004 kg of Helium has 6 x 1023 atoms. So, 431 kg of Helium have 65 x 1028 atoms
1 mole of helium (or 4 g or 0.004 kg) has 6 x 1023 atoms. So, 590 kg of helium will have 8.9 x 1028 atoms
1 mole of helium (or 4 g or 0.004 kg) will have 6 x 1023 atoms. So, 544 kg will have 8.16 x 1028 atoms.
520 kg = 5.20 X 105 grams. The gram atomic mass of helium is 4.0026; therefore the number of moles of helium in 520 kg is (5.20/4.0026) X 105 or 1.30 X 105. Multiplying this number by Avogadro's Number, 6.022 X 1023, yields the number of atoms, which is about 7.82 X 1028, to the justified number of significant digits.
The answer is 50,38349e+23 atoms.
The answer is 6,2729.10e+26 carbon atoms.
to lift 1 kg or 2 pounds you need 0.16 kg of helium so for 2000 pounds you need 160 kg of helium or 320 pounds at 1 atmosphere
113 662.1016 uranium-234 atoms
1.30*1025
12,4439 kg of gold contain 63,177 moles.