5,812 561 184.1024
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
There are 3.80 x 10^24 molecules of CO2 in 6.30 mol. This can be calculated by using Avogadro's number, which is 6.022 x 10^23 molecules/mol.
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.
The number of molecules is 7,2265690284.10e23.
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.
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
There are 1.28x10^24 molecules of SF4. 2.13 mol * 6.022x10^23 molecules/mol = 1.28x10^24 molecules.
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
2C(s) + 2H2(g) + 52.5 kJ -> C2H4
The number of bromine molecules present in the flask can be calculated using Avogadro's number, which is 6.022 x 10^23 molecules/mol. In this case, there are 0.380 mol of bromine, so the number of bromine molecules present is 0.380 mol x 6.022 x 10^23 molecules/mol.
There are (5.41 \times 10^{23}) molecules of (O_2) in 0.900 moles.
1 mole gas = 22.4L 1.5mol C2H4 x 22.4L/mol = 33.6L ethane gas (C2H4)
2C(s) + 2H2(g) + 52.5 kJ -> C2H4
To react with 0.373 mol of ethylene (C2H4), you need an equal number of moles of water (H2O) based on the balanced chemical equation: C2H4 + H2O → C2H5OH This means you need 0.373 mol of water to react with 0.373 mol of ethylene. To convert moles to grams, you would multiply 0.373 mol by the molar mass of water (18.015 g/mol) to get the grams needed.
To find the number of moles in 1.21 molecules of HBr, divide the number of molecules by Avogadro's number (6.022 x 10^23 molecules/mol). Thus, 1.21 molecules of HBr is approximately 2.01 x 10^-24 moles.