Question

Определите массу этана, в которой содержится 12,04∙1023молекул.

Answer 1

N(c2h6)= N/Nª= (12,04*10^23)/(6,02*10^23)=2 моль
m(c2h6)=n*M=2*30=60 г
Ответ: m(c2h6)=60 г

Answer 2

Calculating the Mass of Ethane Containing 12.04*10^23 Molecules

To calculate the mass of ethane containing 12.04*10^23 molecules, we can use the molar mass of ethane and Avogadro's number to find the mass.

Step 1: Find the Molar Mass of Ethane The molar mass of ethane (C2H6) can be calculated by adding the atomic masses of its constituent atoms.

Step 2: Use Avogadro's Number to Find the Mass Once we have the molar mass of ethane, we can use Avogadro's number to find the mass of 12.04*10^23 molecules of ethane.

Calculation

The molar mass of ethane (C2H6) is 30.07 g/mol. Using Avogadro's number (6.022*10^23 mol^-1), we can calculate the mass of 12.04*10^23 molecules of ethane.

Mass of 1 mole of ethane (C2H6): 30.07 g Number of molecules in 1 mole (Avogadro's number): 6.022*10^23 molecules/mol

Now, to find the mass of 12.04*10^23 molecules of ethane: ``` Mass = (Molar mass of ethane) * (Number of molecules) = 30.07 g/mol * 12.04*10^23 molecules ≈ 362.45 g ```

Conclusion

The mass of ethane containing 12.04*10^23 molecules is approximately 362.45 grams.