1. Given: mass = 0.334g of C2H6, CAtomicWeight = 12.01, HAtomicWeight = 1.008
Required:
a. Molar Mass of Ethane
b. Number of molecules of C2H6 in 0.334g of C2H6
Formula:
a. (n = (2 x CAtomicWeight) + (6 x HAtomicWeight)
b. mass x (1 mole / a ) x (6.022x1023 molecules / 1 mole)
Solution:
a. Molar Mass of Ethane
n = (2 x 12.01) + (6 x 1.008)
n = 24.02 + 6.0498
n = 30.0698
b. Number of molecules
n = 0.334 x (1 mole / 30.0698 g) x (6.022x1023 molecules / 1 mole)
n = 6.689x1021 molecules
FINAL Answer: The are 6.689x1021 molecules in 0.334g of Ethane
No, there are many other nonpolar molecules besides carbon dioxide and carbon tetrachloride. Examples include methane, ethane, and diatomic gases like nitrogen and oxygen. Nonpolar molecules generally have more symmetric structures, with balanced distribution of electrons.
12 molecules of carbon 24 of hydrogen 12 of oxygen
To find the number of molecules in 42.0g of Cl2, you first need to determine the number of moles using the molar mass of Cl2 (71 g/mol). Next, you can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules. Therefore, in 42.0g of Cl2, there would be approximately 3.56 x 10^23 molecules.
0.175 X Avogadro's Number = about 1.05 X 1023.
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.
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.
Ethane does not have any molecule of carbon dioxide. However when ethane undergoes combustion then two molecules of carbon dioxide are formed (as ethane contains two carbon atoms).
at STP 1 mole occupies 22.4 litres. 64.28 / 22.4 is 2.8696428 moles. Multiply this by avagadro's constant (6.022*10^23) gives 1.7281x10^24 molecules
To determine the number of molecules in 60 g of ethane, you need to first calculate the number of moles of ethane using its molar mass (30 g/mol). Then, you can convert moles to molecules using Avogadro's number, which is 6.022 x 10^23 molecules/mol.
I am presuming that you are asking how many carbon dioxide (CO2) molecules will be formed when two ethane (C2H6) molecules burn in a plentiful supply of oxygen. The following balanced equation is that of the burning of ethane in a plentiful supply of oxygen: C2H6 + 3.5O2 -----> 2CO2 + 3H2O. The number before each of the molecules in the equation tells us how many there are in this reaction. This means that for every one ethane molecule that is burnt, two carbon dioxide molecules are produced.
You think probable to polymerization.
30 g of ethane will have 6.023 x 1023 molecules of ethane So, 5.5 g will have 1.104 x 1023 molecules of ethane Since there are two carbon atoms, in one molecule of ethane, 5.5 g of ethane will have 2.208 x 1023 atoms of carbon.
The reaction is :- 2C2H6 + 7O2 ----------> 4CO2 + 6H2O When one mole ethane is combusted 7/2 moles of oxygen are used. When 3 moles of ethane are combusted 3 x 7/2 moles of oxygen used. No. of oxygen molecules consumed =6.022 x 1023 x7/2= 21.077 x 1023=2.107 x 1024 molecules.
Only sigma bonds are present in ethane. There is one carbon-carbon sigma bond and six carbon-hydrogen sigma bonds in C2H6.
To calculate the number of molecules in 15 grams of ethane (C2H6), first find the molar mass of C2H6, which is 30.07 g/mol. Next, calculate the number of moles in 15 grams using the formula: moles = mass / molar mass. Finally, use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules.
The chemical formula for ethane is C2H6 and there is 6.022 * 10^23 molecules in every mole. So to find your answer simply multiply 6.022 * 10^23 by 51.2 which gives: 3.083 X 10^25
There is a total of 8 atoms in Ethane.