Draw the structural formula of (3e,5e)-2,5-dibromo-3,5-octadiene.

Chemistry

1 answer:

mrs_skeptik [129]3 years ago

4 0

We need draw the structure of (3E,5E)-2,5-dibromo-3,5-octadiene.

The structure is shown below:

The molecule has two double bonds in 3 and 5 position and two Br-atoms attached to 2 and 5 position.

The configuration is 3E and 5E. E configuration indicates cis- isomers where similar groups are in same side with respect to double bonds.

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In the reaction Mg (s) + 2HCl (aq) -> H2 (g) + MgCl2 (aq), how many grams of hydrogen gas will be produced from 300 millilite

d1i1m1o1n [39]

Mg(s) + 2HCl(aq) = H₂(g) + MgCl₂(aq)

v=300 ml=0.3 l
c=4 mol/l

n(HCl)=vc

m(H₂)/M(H₂)=n(HCl)/2

m(H₂)=M(H₂)vc/2

m(H₂)=2.0g/mol·0.3 l · 4mol/l /2 = 1.2 g

1.2 grams of hydrogen gas will be produced

5 0

4 years ago

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

algol13

Answer:

1) C₃H₄N

Explanation:

The empirical formula of a compound is the formula that gives the positive integer ratio of the atoms of the elements in the compound in the simplest form

The mass of carbon in the compound = 90 grams

The molar mass of carbon = 12.011 g/mol

The number of moles of carbon = 90 g/(12.011 g/mol) ≈ 7.4931313 moles

The mass of hydrogen in the compound = 11 grams

The molar mass of hydrogen = 1.00794 g/mol

The number of moles of carbon = 11 g/(1.00794 g/mol) ≈ 10.913348 moles

The mass of nitrogen in the compound = 35 grams

The molar mass of nitrogen = 14.0067 g/mol

The number of moles of carbon = 35 g/(14.0067 g/mol) ≈ 2.49880414 moles

Dividing by the smallest mole ratio gives;

The proportion of carbon, C = 7.4931313/2.49880414 = 2.9987 ≈ 3

The proportion of nitrogen, N = 10.913348 /2.49880414 = 4.367 ≈ 4

The proportion of nitrogen, N = 2.49880414 /2.49880414 = 1

Therefore, the empirical formula of the compound is C₃H₄N.

7 0

3 years ago

N2 (g) + 3 H2 (g) → 2 NH3 (g)

shusha [124]

This problem is asking for the rate of disappearance of gaseous nitrogen, given the rate of appearance of ammonia and the chemical reaction. At the end, the result turns out to be -0.228 M/s.

<h3>Rates of appearance and disappearance</h3>

In chemical kinetics, one of the most relevant calculations are based on rates of appearance and disappearance of chemical species in a chemical reaction. This can be calculated via rate portions based on the stoichiometric coefficients in the reaction.

Thus, for this problem, one can write:

$\frac{r_{N_2}}{-1} =\frac{r_{H_2}}{-3} =\frac{r_{NH_3}}{2}$

Where the rate of appearance or disappearance is divided by the stoichiometric coefficient. Therefore, one can solve for the rate of disappearance of N2 with:

$\frac{r_{N_2}}{-1} =\frac{r_{NH_3}}{2}\\\\r_{N_2}=\frac{r_{NH_3}*-1}{2}\\\\r_{N_2}=\frac{0.456M/s*-1}{2}\\\\r_{N_2}=-0.228M/s$