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oksano4ka [1.4K]
3 years ago
14

Please help this test ends in 30 min

Chemistry
1 answer:
RideAnS [48]3 years ago
8 0

Answer:

Option C

CH₃CH₂CH₂COOH

Explanation:

Carbonxylic acids are compounds which has the general formula

R–COOH where R is an alkyl group.

Considering the options given in the question above,

For A:

CH₃CH₂OCH₂CH₃ is an ether compound with general formula ROR' where R and R' are both alkyl group.

For B:

CH₃CH₂CH₂CH₂OH is an alcohol with general formula ROH where R is an alkyl group.

For C:

CH₃CH₂CH₂COOH is a carbonxylic acid with general formula R–COOH where R is an alkyl group.

For D:

CH₃CH₂C=OCH₂CH₃ is a ketone compound with general formula RC=OR' where R and R' are both alkyl group

For E:

ClCH₂CH₂CH₂CH₂CH₂CH₂Br is simply an Alkyl halide with general formula XRX where X is an halogen (i.e F, Cl, Br or I) and R is an alkyl group.

From the above illustration, only option C contains a Carbonxylic compound.

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The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO2 and 3.969 mg of
attashe74 [19]

Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms  of C , H , O in the compound

= 2.64 / 12 : .441 /1 : .714 / 16

= .22 : .44 : .0446

= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

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= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

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Let molecular formula be (C₅H₁₀O)ₙ

molecular weight = n ( 5 x 12 + 10 + 16 )

= 86 n

86 n = 88.6

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So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

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