Chemistry, 09.09.2020 14:01 mirandahamamaox89fl

air is bubbled through a drum of liquid hexane at a rate of 0.200 kmol / min. the gas stream leaving the drum contains 10.0 mole's hexane vapor. air may be considered insoluble in liquid hexane. use an integral balance to estirmate the time reguired to vaporize in min to vaporize 10.0 m of the liquid. if the total moles of the products 0.222 kmol / min and the specific gravity of liquid hexane is 0.659 and the accumulation term equals -76.45 kmol ch4

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Answer from: doll1234

Complete Question

Air is bubbled through a drum of liquid hexane at a rate of 0.200 kmol / min. the gas stream leaving the drum contains 10.0% mole's hexane vapor. air may be considered insoluble in liquid hexane. use an integral balance to estirmate the time reguired to vaporize in min to vaporize 10.0 m^3 of the liquid.if the total moles of the products 0.222 kmol / min and the specific gravity of liquid hexane is 0.659 and the accumulation term equals -76.45 kmol ch4

The value is $t = 3851.7 \ min$

Explanation:

From the question we are told that

The rate at which the air is bubbled is $R = 0.200 \ kmol/min$

The rate at which the gas stream is leaving the drum is $r = 0.222 \ kmol/min$

The percentage of hexane contained in the gas leaving the drum $10\%$

The volume of liquid hexane to be vaporized is $V = 10 \ m^3$

The specific gravity of liquid hexane is $SG_{H} = 0.659$

Generally the density of the liquid is mathematically evaluated as

$\rho_h = \rho_w * SG_{H}$

Here $\rho_w$ is the density of water with value

$\rho_w = 1000 \ kg/m^3$

$\rho_h = 1000* 0.659$

$\rho_h =659$

The mass of the liquid is mathematically represented as

$m = \rho_h * V$

m = 659 * 10V

$m = 6590 \ kg$

The number of moles of liquid hexane is mathematically represented as

$n = \frac{m}{ M}$

Here M is the molar mass of hexane with value $M = 0.086 \ kg$

$n = \frac{6590 }{ 0.086}$

n = 76628 mols

$n = 76.628 \ k mols$

The rate at which hexane liquid is vaporized is mathematically represented as

$z = 10\% * r$

z = 0.1 * 0.222

$z = 0.0222 \ kmol/min$

Generally the time required to vaporize the the liquid(i.e n) is mathematically represented as

$t = \frac{n}{z}$

=> $t = \frac{76.628}{0.0222}$

=> $t = 3851.7 \ min$

Answer from: Quest

Mass is the amount of matter in an object. weight is the measurement of the pull of gravity on an object. say for example, you travel to the moon which has a different gravitational field compared to the gravitational field of earth. (gravitational fields: earth = 1, moon = 0.165) now when you are on the moon, the pull of gravity on you is different than on the earth. and as weight is the measurement of the pull of gravity on an object, that means yes, your weight can and will change on the moon. however, on the moon, on earth or even on jupiter, the amount of mass inside of you remains the same. meaning your physical mass remains constant. in conclusion it is possible for your weight to change as your mass remains constant.

Answer from: Quest

Use a magnet to separate iron, evaporate water, i'm not sure how to separate the other two i'm sorry but good luck

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