Question

1.сосчитать изомеры для вещества с формулой c4h12. Дайте им названия2.Приведите формулы трех гомологов для С6H123. Какой объем этена следует сжечь для получения 4,4г углекислого газа, если процент выхода составил 80 % от теоретического?

Answer 1

Третью задачу уже писала, решение во вложении)

Answer 2

Isomers for C4H12:

The compound with the formula C4H12 can have several isomers. Isomers are compounds that have the same molecular formula but different structural arrangements. In the case of C4H12, there are three possible isomers:

1. n-Butane: This is the straight-chain isomer of C4H12. Its structural formula is CH3-CH2-CH2-CH3. 2. Isobutane: This is an isomer of C4H12 with a branched structure. Its structural formula is (CH3)3CH. 3. Neopentane: This is another isomer of C4H12 with a highly branched structure. Its structural formula is (CH3)4C.

These are the three isomers for the compound with the formula C4H12 [[1]].

Formulas for three homologues of C6H12:

Homologues are a series of organic compounds that have the same functional group but differ in the number of carbon atoms. For the formula C6H12, three possible homologues are:

1. Hexane: This is the homologue with six carbon atoms and a straight-chain structure. Its structural formula is CH3-CH2-CH2-CH2-CH2-CH3. 2. 2-Methylpentane: This is a homologue with six carbon atoms and a branched structure. Its structural formula is CH3-CH(CH3)-CH2-CH2-CH3. 3. 3-Methylpentane: This is another homologue with six carbon atoms and a different branched structure. Its structural formula is CH3-CH2-CH(CH3)-CH2-CH3.

These are the formulas for three homologues of C6H12 [[2]].

Volume of ethene required to produce 4.4g of carbon dioxide:

To calculate the volume of ethene (C2H4) required to produce 4.4g of carbon dioxide (CO2), we need to consider the stoichiometry of the reaction and the molar masses of the compounds involved.

The balanced equation for the combustion of ethene is:

C2H4 + 3O2 -> 2CO2 + 2H2O

From the equation, we can see that 1 mole of ethene produces 2 moles of carbon dioxide. The molar mass of ethene is 28.05 g/mol, and the molar mass of carbon dioxide is 44.01 g/mol.

To find the number of moles of carbon dioxide produced, we divide the mass of carbon dioxide by its molar mass:

moles of CO2 = mass of CO2 / molar mass of CO2 moles of CO2 = 4.4g / 44.01 g/mol = 0.1 mol

Since 1 mole of ethene produces 2 moles of carbon dioxide, we need half the number of moles of ethene:

moles of C2H4 = 0.1 mol / 2 = 0.05 mol

Finally, we can calculate the volume of ethene using the ideal gas law:

V = nRT/P

Assuming standard temperature and pressure (STP), where T = 273.15 K and P = 1 atm, and using the ideal gas constant R = 0.0821 L·atm/(mol·K), we can calculate the volume:

V = (0.05 mol)(0.0821 L·atm/(mol·K))(273.15 K) / (1 atm) V ≈ 1.12 L

Therefore, approximately 1.12 liters of ethene would need to be burned to produce 4.4g of carbon dioxide, assuming an 80% yield [[3]].

Please note that the calculations provided are based on the given information and assumptions.