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Burka [1]
2 years ago
11

If 1,2-dichloroethane has a density of 1.23 g/ml, how do you expect it to sit in a separatory funnel with an aqueous solution?

Chemistry
1 answer:
Jobisdone [24]2 years ago
5 0

1,2-dichloroethane of density 1.23 g/ml would settle down an aqueous solution.

<h3>Density:</h3>

Knowing whether the aqueous layer is above or below the organic layer in the separatory funnel is crucial since it determines which layer is ultimately kept and discarded. Because immiscible solvents have different densities, they will stack on top of one another. The less-dense solution will rest on top, and the more dense one will rest at the bottom.

Due to their low density of less than 1 g/mL, the majority of non-halogenated organic solvents will float on top of an aqueous solution (if they are immiscible). One significant exception is that halogenated solvents will sink below aqueous solutions because they are denser than water (having densities greater than 1 g/mL). Except for halogenated solvents like dichloromethane, which are usually on the bottom, most organic solvents, such as diethyl ether, are on top.

Learn more about aqueous layer here:

brainly.com/question/14356327

#SPJ4

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
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