1.8x10*24
1 mole = 6.022e23 atoms 8.25 mole = 4.96815 e 24 atoms
This is equivalent to 1,4 moles.
1,0.10e9 atoms is equivalent to 0,166.10e-14 moles.
Seventeen moles of hydrogen correspond to 204,75.10e23 atoms.
2,50 moles of silicon contain 15,055352142.10e23 atoms.
To find the number of moles of Cr in the sample, first calculate the molar mass of Cr (chromium). Chromium has an atomic mass of approximately 51.996 g/mol. Next, divide the number of atoms by Avogadro's number (6.022 x 10^23 atoms/mol) to convert atoms to moles. So, 4.37 x 10^23 atoms of Cr is equivalent to 0.725 moles of chromium.
The balanced chemical equation for this reaction is: Cr2O3 + 3H2 -> 2Cr + 3H2O. So, according to the equation, 3 moles of hydrogen are needed to convert 1 mole of chromium oxide. Therefore, to convert 5 moles of chromium oxide, 15 moles of hydrogen would be needed.
1 mole = 6.022e23 atoms 8.25 mole = 4.96815 e 24 atoms
To find the number of moles in 3.6 x 10^24 atoms of chromium, divide the number of atoms by Avogadro's number (6.022 x 10^23 atoms/mol). 3.6 x 10^24 atoms of Cr / 6.022 x 10^23 atoms/mol ≈ 5.98 moles of Cr.
This is some easy math: chromium (Cr, #24) has an atomic mass of 52 grams per mole (from the periodic table). According to the Avogadro constant, 1 mole of any substance contains 6.02x1023 representative particles. So, 52g of Cr is 1 mole, and contains 6.02x1023 atoms.
1,638 moles contain 9,864266723766.10e23 atoms.
To determine how many miles of calcium nitrate would react with 4.55 moles of Chromium (III) sulfate to produce chromium (III) nitrate, you would first need to write the balanced chemical equation for the reaction between calcium nitrate and chromium (III) sulfate. Then, use the stoichiometry of the reaction to convert moles of Chromium (III) sulfate to moles of calcium nitrate, and finally, convert moles of calcium nitrate to miles using the molar mass.
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.
20 moles
This is equivalent to 1,4 moles.
There are twice as many oxygen atoms as carbon atoms in carbon dioxide, so 100.0 moles of carbon dioxide would contain 200.0 moles of oxygen atoms.
1,0.10e9 atoms is equivalent to 0,166.10e-14 moles.