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LiRa [457]
3 years ago
10

Which of the following soils is the most porous? Soil A - flat, sorted particles. Soil B - round, sorted particles Soil C - roun

d, unsorted particles Soil D - flat, unsorted particles
Chemistry
1 answer:
Ksju [112]3 years ago
4 0

Answer: Option (b) is the correct answer.

Explanation:

Porosity means the ease with which a fluid can pass through between a rock or cracks etc.

If particles of a soil are flat and even if they are sorted then there will rarely be any space available for fluid or water to pass through it.

If soil particles are round in shape and are sorted then there will be space available for the fluid to pass through it. Hence, this type of soil will be most porous.

Similarly, if soil particles are round but are unsorted then there will be less space available as compared to sorted particles. This is because particles are placed randomly so, somewhere there will be much less space and somewhere there will be large space available. As a result, soil will not be most porous.

If soil particles are flat and unsorted then there will also be less space available for the fluid to pass through it. Thus, soil will be less porous.

Hence, we can conclude that out of the given options, soil B - round, sorted particles is the most porous.

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You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

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