To find the number of helium atoms in a helium blimp, you would first need to convert the mass of helium (431 kg) to moles using the molar mass of helium. Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to calculate the number of helium atoms.
To find the number of helium atoms, we need to convert the mass of helium to moles and then use Avogadro's number to convert moles to atoms. The molar mass of helium is 4 g/mol. First, convert 590 kg to grams (590,000 g). Then, divide by the molar mass of helium to find moles, and finally multiply by Avogadro's number (6.022 x 10^23 atoms/mol) to get the number of atoms.
The answer is 50,38349e+23 atoms.
12,4439 kg of gold contain 63,177 moles.
One pound is 0.4536 kg. 44g is the molar mass of CO2, so: 44 g contain the Avogadro number of carbon atoms, 6.023 x 10^23 so 1 g contains 6.023/44 x 10^23 atoms 1 kg contains 6.023/44 x 10^26 atoms 0.4536 kg contain 6.023 x 0.4536/44 x 10^26 atoms = 6.209 x 10^24 atoms.
To find the number of helium atoms in the blimp, you first calculate the number of moles of helium in 533 kg of helium using the molar mass of helium. Then, you use Avogadro's number (6.022 x 10^23) to convert moles to atoms. The final answer will give you the number of helium atoms in the blimp.
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.
To find the number of helium atoms in 542 kg of helium, you need to first calculate the number of moles of helium in 542 kg using the molar mass of helium, which is 4 grams/mol. Then, you can use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles of helium to atoms.
To find the number of helium atoms in a helium blimp, you would first need to convert the mass of helium (431 kg) to moles using the molar mass of helium. Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to calculate the number of helium atoms.
To find the number of helium atoms, we need to convert the mass of helium to moles and then use Avogadro's number to convert moles to atoms. The molar mass of helium is 4 g/mol. First, convert 590 kg to grams (590,000 g). Then, divide by the molar mass of helium to find moles, and finally multiply by Avogadro's number (6.022 x 10^23 atoms/mol) to get the number of 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.
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
12,4439 kg of gold contain 63,177 moles.
There are approximately 1.19 x 10^28 atoms of U-234 in 1000 kg of natural uranium.
There are approximately 1.04 x 10^25 carbon atoms in a 12.5 kg sample of carbon. This can be calculated using Avogadro's number and the molar mass of carbon.
Quite a few! 147.6 kg chromium (1000 grams/1 kg)(1 mole Cr/52.00 grams)(6.022 X 1023/1 mole Cr) = 1.709 X 1027 atoms of chromium =========================