5,812 561 184.1024
There are approximately 5.8 x 10^24 molecules in 9.6 mol of C2H4. This is calculated using Avogadro's number, which is 6.022 x 10^23 molecules/mol.
To determine the number of molecules in 45.8 mg of C2H4, we first calculate the number of moles using the molar mass of C2H4 (28.05 g/mol). Then we use Avogadro's number (6.022 x 10^23 molecules/mol) to find the number of molecules, which is approximately 1.23 x 10^20 molecules.
2C(s) + 2H2(g) + 52.5 kJ -> C2H4
1296 g N2O5 x 1 mol N2O5/108 g x 6.02x10^23 molecules/mole = 7.22x10^24 molecules
The number of molecules is 7,2265690284.10e23.
my Glizzy
C2H4 + H2O --> C2H5OHReaction balanced at 1:1:1 mole of each compound, so you'll need 0.132 mol C2H4 and this is equal to:0.132 (mol C2H4) * 28 (g/mol C2H4) = 3.696 g C2H4 = 3.70 g C2H4
2.01 mol
There are 1.28x10^24 molecules of SF4. 2.13 mol * 6.022x10^23 molecules/mol = 1.28x10^24 molecules.
1 mole gas = 22.4L 1.5mol C2H4 x 22.4L/mol = 33.6L ethane gas (C2H4)
C2H4 + 3 O2 --> 2 CO2 + 2 H2OSo 2.16 mol O2 will produce 1.44 mol H2O(and 1.44 mol CO2)because 3:2 = 2.16 : 1.44
2.65 mol * 64.07 g/mol = 169.79 g
5.418E23 molecules
? molc' MgBr2 = (13.77lbs)(453.59g/1lbs)(1 mol/184.11g)(6.02e23 molc'/1 mol) 13.77 lbs =2.04e25 molecules
4.91 mol * 6.02214129(27)×1023 / mol = 2.96 ×1024
There are 6.022x10^23 molecules in 1.00 mol of anything.
To find the number of molecules of ethane in 0.334 grams, you would first convert the mass to moles using the molar mass of ethane. Then, you can use Avogadro's number (6.022 x 10^23 molecules/mol) to find the number of molecules in that number of moles.
2C(s) + 2H2(g) + 52.5 kJ -> C2H4