All categories

2 years ago

How will you separate a solution of benzene and chloroform?

Chemistry

1 answer:

Troyanec [42]2 years ago

7 0

Answer:

$\huge\boxed{\sf Distillation}$

Explanation:

State of benzene at RTP = liquid
State of chloroform at RTP = liquid
Boiling point of benzene = 80.1 °C
Boiling point of chloroform = 61.2 °C

Since, both of the chemicals are liquids, we can separate it by the process of distillation.

<u>Distillation:</u>

is the process in which we separate two liquids on the basis of their difference in boiling points.

<u>How it works:</u>

Since chloroform has less boiling point, it will evaporate and collected first and benzene will follow it after sometime.

- Apparatus of distillation is in the attached file.

$\rule[225]{225}{2}$

You might be interested in

Radiation, Chemical agents and Viruses.

zmey [24]

Radiation, Chemical agents and viruses may all cause C. Generic Mutations.

7 0

3 years ago

Read 2 more answers

What is true when an element is oxidized? It bonds with the hydroxide ion. It loses electrons to another element. It reacts with

Mama L [17]

Answer: It loses electrons to another element.

Explanation:- Oxidation is the process in which an element loses electrons and there is an increase in the oxidation state. On losing electrons it combines with a electronegative element such as oxygen, sulphur or nitrogen etc.

$Fe\rightarrow Fe^{2+}+2e^-$

Reduction is the process in which an element gains electrons and there is a decrease in the oxidation state.

$\frac{1}{2}O_2+2e^-\rightarrow O^{2-}$

8 0

3 years ago

How many Liters are in 1.245 x 10 6mL?

Kamila [148]

1245 liters because you move the decimal place over to the right six times

4 0

3 years ago

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i

alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

Explanation:

The pH of the solution = 2.46

$pH=-\log[H^+]$

$2.46=-\log[H^+]$

$[H^+]=0.003467 M$

$HA\rightleftharpoons H^++A^-$

Initially

0.0144 0 0

At equilibrium

(0.0144-x) x x

The expression if an dissociation constant is given by :

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(0.0144-x)}$

$x=[H^+]=0.003467 M$

$K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}$

$K_a=1.099\times 10^{-3}$

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

3 0

3 years ago

How many degrees of unsaturation are lost or gained in the base-induced transformation from mito-ph to its photoactive ring-open

Fudgin [204]

Unsaturation (IHD) 2 hydrogen Needed

IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)

Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2

The degrees of unsaturation in a molecule are additive — a molecule with one double bond has one degree of unsaturation, a molecule with two double bonds has two degrees of unsaturation, and so forth.

6 0

3 years ago