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1 year ago

__________ could be the formula of an alkene. a. c3h8 b. c2h8 c. c6h6 d. c3h6 e. c17h36

1 answer:

6 0

The general formula for alkenes is CnH2n, the formula with hydrogen count double the carbon count should be the correct formula for alkene and that is d. C3H6.

What are alkenes?

Alkenes, commonly known as olefins, are organic unsaturated hydrocarbons that have one or more carbon-carbon double bonds in their chemical structure and are composed of carbon and hydrogen atoms.

Alkenes are unsaturated hydrocarbons with a double bond between the carbon atoms. Carbon atoms are connected by at least one double bond. The general formula for alkenes is C n H 2n. Olefin is frequently substituted with alkenes. The word "olefin" comes from the Greek phrase "olefin gas," which denotes the production of oil.

<u>Since, their general formula is CnH2n, the formula with hydrogen count double the carbon count should be the correct formula for alkene and that is d. C3H6.</u>

To learn more about alkenes from the given link below,

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Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz

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Answer: Freezing point of a solution will be $-1.16^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

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i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant = $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

$(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}$

$(5.50-T_f)^0C=6.68$

$T_f=-1.16^0C$

Thus the freezing point of a solution will be $-1.16^0C$

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