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dedylja [7]
2 years ago
10

Chemistry help!!

Chemistry
1 answer:
Scrat [10]2 years ago
8 0

The student originally has 252 grams of water in this experiment.

LAW OF CONSERVATION OF MASS:

  • The law of conservation of mass explains that matter (mass) can neither be created nor destroyed but can be changed from one form to another.

  • This means that in a chemical reaction, the sum of the masses of the reactants must equate to the total mass of product(s).

  • According to this question, a student conducts an experiment to separate water into hydrogen and oxygen. The student collects 28.0 g of hydrogen and 224.0 g of oxygen.

  • Since hydrogen and oxygen are the constituent elements of water, the sum of their masses must equate the mass of water.

  • Therefore, 224g of oxygen + 28g of hydrogen = 252g of water.

Learn more at: brainly.com/question/24996173?referrer=searchResults

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Exposing a 250 mL sample of water at 20.∘C to an atmosphere containing a gaseous solute at 27.59 kPa resulted in the dissolution
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Answer : The solubility of this gaseous solute will be, 5.18\times 10^{-5}g/mL

Explanation :

First we have to calculate the concentration of solute.

\text{Concentration of solute}=\frac{Mass}{Volume}=\frac{4.51\times 10^{-3}g}{250mL}=1.80\times 10^{-5}g/mL

Now we have to calculate the Henry's law constant.

Using Henry's law :

C=k_H\times p

where,

C = concentration of solute = 1.80\times 10^{-5}g/mL

p = partial pressure = 27.59 kPa

k_H = Henry's law constant = ?

Now put all the given values in the above formula, we get:

1.80\times 10^{-5}g/mL=k_H\times (27.59kPa)

K_H=6.52\times 10^{-7}g/mL.kPa

Now we have to calculate the solubility of this gaseous solute when its pressure is 79.39 kPa.

C_{(79.39kPa)}=k_H\times p

C_{(79.39kPa)}=(6.52\times 10^{-7}g/mL.kPa)\times (79.39kPa)

C_{(79.39kPa)}=5.18\times 10^{-5}g/mL

Therefore, the solubility of this gaseous solute will be, 5.18\times 10^{-5}g/mL

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The choices are:
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