Question

3, 5 грамм алкена присоединяют 20 грамм Br. Определить молекулярную формулу вещества. Подробное решение.Срочно!

Answer 1

1) СnH2n + Br2 = CnH2n + Br2
2) По уравнению: n(CnH2n) = n(Br2)
3) n(Br2) = m(Br2)/M(Br2) = 20/160 = 0.125 моль 
4) n(CnH2n) = n(Br2) = 0.125 моль 
5) М(CnH2n) = m(CnH2n)/n(CnH2n) = 3,5/0.125 = 28 г/моль 
6) 14n=28
    n=2
7) C2H4 (этен)
Ответ: C2H4

Answer 2

СnH2n+Br2=CnH2nBr2
одномолярные отношения
1)n(Br2)=20/160=0.125моль
n(CnH2n)=n(Br2)=0.125моль
2)М(CnH2n)=3,5/0.125=28г/моль
14n=28
n=2
C2H4-этен

Answer 3

Calculation of Molecular Formula

To determine the molecular formula of the substance formed by the addition of 3 grams of alkene and 20 grams of Br, we need to consider the molar masses of the elements involved.

The molar mass of bromine (Br) is approximately 79.9 g/mol.

Since we know the mass of alkene (3 grams) and bromine (20 grams), we can calculate the number of moles of each element using their respective molar masses.

The number of moles of alkene can be calculated using the formula:

moles = mass / molar mass

For the alkene, the number of moles is:

moles of alkene = 3 g / molar mass of alkene

Unfortunately, the molar mass of the alkene is not provided in the question, so we cannot calculate the exact number of moles of alkene.

However, we can still determine the molecular formula of the substance by assuming a molar mass for the alkene and using the given ratio of 3 grams of alkene to 20 grams of bromine.

Let's assume a molar mass of 42 g/mol for the alkene. This is just an example value for demonstration purposes.

Using this assumed molar mass, we can calculate the number of moles of alkene:

moles of alkene = 3 g / 42 g/mol

Now, let's calculate the number of moles of bromine:

moles of bromine = 20 g / 79.9 g/mol

To find the empirical formula, we need to determine the ratio of moles between the alkene and bromine.

Dividing the moles of alkene by the smallest number of moles obtained, we get:

moles of alkene / smallest moles = ratio of moles

Let's assume that the moles of alkene obtained is 0.0714 and the moles of bromine obtained is 0.2503. Dividing the moles of alkene by the smallest moles (0.0714 / 0.0714), we get a ratio of 1:3.

Therefore, the empirical formula of the substance is C1H3Br3.

However, without the exact molar mass of the alkene, we cannot determine the molecular formula of the substance. The molecular formula could be different depending on the actual molar mass of the alkene.

Please note that the values used in this example are hypothetical and for demonstration purposes only. The actual molecular formula may vary depending on the specific alkene used.

I hope this explanation helps! Let me know if you have any further questions.