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3 years ago

6

Почему молекулы алканов в пространстве имеют зигзагообразную форму

1 answer:

Solnce55 [7]3 years ago

6 0

Answer:

Alkanes (also saturated hydrocarbons , paraffins ) are acyclic hydrocarbons of a linear or branched structure, containing only simple bonds and forming a homological series with the general formula C n H 2n + 2 .

Explanation:

ll alkanes belong to a larger class of aliphatic hydrocarbons. Alkanes are saturated hydrocarbons, that is, they contain the maximum possible number of hydrogen atoms for a given number of carbon atoms. Each carbon atom in the alkane molecules is in a state of sp 3 - hybridization - all 4 hybrid orbitals of the C atom are identical in shape and energy, 4 bonds are directed to the vertices of the tetrahedron at angles of 109 ° 28 '. C – C bonds are σ bonds characterized by low polarity and polarizability . The C – C bond length is 0.154 nm , the C – H bond length is 0.1087 nm.

The simplest member of the class is methane (CH 4 ). The hydrocarbon with the longest chain — noncontactrictan C 390 H 782 — was synthesized in 1985 by English chemists I. Bidd and MK K. Whiting

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Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

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Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

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Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

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