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2 years ago

why is often necessary to follow extensive purification procedures in the isolation of organic compounds?

Chemistry

1 answer:

GREYUIT [131]2 years ago

6 0

To remove impurities

Explanation:

There are different substances present in an organic compound. If an organic compound contains different substances then there are some physical or chemical methods for the process of isolation, so that we will get different substances from the organic compound, which may or may not contain impurities.
To get rid of impurities after the isolation process purification process has to be done, so that we will get a purified substance from an organic compound.
Purification may be done using the method of sublimation, distillation and so on.

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3 years ago

For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the

Radda [10]

Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).
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1/T₁ = 1/622 K = 0,0016 1/K.
1/T₂ = 1/666 K = 0,0015 1/K.
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2 years ago

A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while th

sleet_krkn [62]

Answer:

$p_{H_2O}=2.00atm$

Explanation:

Hello!

In this case, according to the following chemical reaction:

$2H_2+O_2\rightarrow 2H_2O$

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

$n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2$

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

$n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O$

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

$p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm$

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3 years ago

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Answer: D

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3 years ago

Read 2 more answers

Analysis of an ore of calcium shows that it contains 13.61 g calcium and 21.77 g oxygen in a 46.28-g sample. Calculate the perce

kondor19780726 [428]

Answer:

29.41% of Calcium and 47.04% of Oxygen

Explanation:

The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.

The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:

13.61g / 46.28g * 100 = 29.41% of Calcium

And percent composition of Oxygen is:

21.77g / 46.28g * 100 = 47.04% of Oxygen

5 0

2 years ago

why is often necessary to follow extensive purification procedures in the isolation of organic compounds?​

why is often necessary to follow extensive purification procedures in the isolation of organic compounds?