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Ipatiy [6.2K]
3 years ago
12

150 lb of NaCl are initially dissolved in 500 gal of a NaCl solution in a container. Water is continuously added to the containe

r at a rate of 5 gal/min and the container content, which is kept uniform by mixing, is also withdrawn out of the container at 5 gal/min. Determine how much NaCl will be in the tank at the end of 1.5 hours.
Chemistry
1 answer:
puteri [66]3 years ago
5 0

Answer:

After 90 minutes there will be 60.5 lb NaCl in the tank

Explanation:

Initially, you have 150 lb/500 gal = 0.300 lb/gal NaCl

After 1 minute, the volume will be the same but you will have less NaCl in the solution.

After the first minute, the mass of NaCl you will lose will be:

0.3 lb/gal · 5 gal = 1.5 lb

Then, the concentration of the solution will be:

(150 lb - 1.5 lb)/ 500 gal = 0.297 lb/gal

After the second minute:

mass of NaCl lost = 0.297 lb/gal · 5 gal = 1.485 lb

concentration of the solution = (148.5 lb - 1.485 lb) / 500 gal = 0.29403 lb/gal

After the third minute:

mass of NaCl lost = 0.294 lb/gal · 5 gal = 1.47015 lb

concentration of the solution = 147.015 lb - 1.47015 lb / 500 gal = 0.2910897 lb/gal

Notice that the solution is being diluted by 1% every minute. Then, the concentration of the solution after every minute will be 99% of the previous solution. For example, after the first minute, the concentration of the solution  will be 99% of the original solution (99% of 0.3 lb/gal = 0.297 lb/gal). Mathematically, we can express it as follows:

0.3  lb/gal · 0.99 = 0.297 lb/gal

After the second minute:

0.297 lb/gal · 0.99 = 0.29403 lb/gal

Since 0.297 lb/gal = 0.3  lb/gal · 0.99

we can write for the second minute:

0.3  lb/gal · 0.99 · 0.99 = 0.29403 lb/gal

And for the third minute:

0.3  lb/gal · 0.99 · 0.99 · 0.99 = 0.2910897 lb/gal

And for the nth minute:

0.3 lb/gal · 0.99ⁿ = concentration of the solution after n minutes

Now, we can find the concentration of the solution after 90 min:

0.300 lb/gal · 0.99^90 = 0.121 lb/gal

The mass of NaCl after 90 minutes will be:

0.121 lb/gal · 500 gal = 60.5 lb

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I assume you meant \text{K}_{2}\text{SO}_{4}.

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PLEASE SOMEBODY EXPLAIN THIS :(((
djverab [1.8K]

Answer:

\boxed{ \sf \: R_f  \: value \: of \: sample \: 1 =0.3142}

<h3>Explanation:</h3>

In Analytical Chemistry chromatography is widely used for the separation of samples.

  • In thin layer chromatography, the mixture of components are separated on the basis of their polarity.
  • The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.
  • Consider the mixture we have taken consist of two samples having large polar difference.
  • Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.
  • The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.

In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.

Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.

Formula to calculate Rf value is

\sf R_f  \: value = \frac{distance \: moved \: by \: sample}{distance \: moved \: by \: solvent}

Now, solving for Rf value of sample 1

<em>Given:</em>

Distance moved by sample 1 = 11 mm

Distance movedby solvent = 35 mm

<em>To find:</em>

Rf value of sample 1 = ?

<em>Solution:</em>

Substituting the given data in above formula,

\small \sf R_f  \: value = \frac{distance \: moved \: by \: sample \: 1}{distance \: moved \: by \: solvent}   \\  \small \sf R_f  \: value =  \cancel\frac{11  \: mm}{35 \:  mm}  = 0.3142

\small \boxed{ \sf \: R_f  \: value \: of \: sample \: 1 =0.3142}

<em><u>Thanks for joining brainly community!</u></em>

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