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nlexa [21]
3 years ago
15

A researcher measures the equilibrium separation between two flat hydrophilic mica plates in a solution of non-ionic surfactant.

When the surfactant concentration is below CMC the surfaces touch at 0 nm distance without any separation (i.e., there is no surfactant adsorption and repulsion)
a) The researcher increases the surfactant concentration much above CMC and finds out that the system has equilibrium separations at 0 nm, 4 nm, 8 nm, 12 nm and some at larger distances. Draw schematics and explain the reason for this effect.
b) The researcher hydrophobizes the mica surfaces and again dips them in the first solution with surfactant concentration below CMC. When these surfaces are brought in contact, they remain separated at a distance of 4 nm (no equilibria at larger separations). Draw a schematic and explain the reason for this observation with hydrophobic surfaces.

Chemistry
1 answer:
svet-max [94.6K]3 years ago
3 0

Answer:

See explaination and attachment

Explanation:

please kindly see attachment for the step by step solution of the given problem.

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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
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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.

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Answer:

0.046 %

Explanation:

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The rate-out

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

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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

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