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How many C2H4 molecules are contained in 45.8 mg of C2H4?

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.


How many grams of ethylene are needed to react with 0.132 mol of H20?

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


How many molecules of sulfur tetrafluoride are present in 2.13 moles of this compound?

There are 1.28x10^24 molecules of SF4. 2.13 mol * 6.022x10^23 molecules/mol = 1.28x10^24 molecules.


How many molecules of water can be produced by the reaction of 2.16 mol of oxygen with excess Ethylene.?

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


How many molecules OS CO2 are present in 51.5 of CO2?

2.65 mol * 64.07 g/mol = 169.79 g


Which reaction shows that the enthalpy of formation of C2H4 is Hf 52.5 kJ mol?

2C(s) + 2H2(g) + 52.5 kJ -> C2H4


A flask contains 0.380 mol of liquid bromine Determine the number of bromine molecules present in the flask?

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.


How many Molecules are contained in 0.900 mol O2?

There are (5.41 \times 10^{23}) molecules of (O_2) in 0.900 moles.


What volume will 1.50 mol of C2H4 gas occupy at STP?

1 mole gas = 22.4L 1.5mol C2H4 x 22.4L/mol = 33.6L ethane gas (C2H4)


which reaction shows that the enthalpy of formation of C2H4 is Hf = 52.5 kJ/mol?

2C(s) + 2H2(g) + 52.5 kJ -> C2H4


How many grams of water are needed to react with 0.373 mol of ethylene?

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.


How many moles are present in 1.21 molecules of HBr?

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.