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Art [367]
2 years ago
11

The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 101.3 k

Pa and the vapor pressure of water under the experimental conditions is 2.41 kPa, what is the pressure (kPa) of the H 2 gas
Chemistry
1 answer:
Gnom [1K]2 years ago
3 0

The H2 produced in a chemical reaction is collected through water in a eudiometer. The pressure (kPa) of the H2 gas is 98.89 kPa

The total pressure in a chemical reaction is the total sum of the partial pressure and the vapor pressure of the chemical substances taking place in the chemical reaction.

  • Total pressure = partial pressure of H₂ gas + vapor pressure of H₂O

∴

The vapor pressure of H₂ gas = Total pressure in the eudiometer - partial pressure of H₂O

Given that:

  • The total pressure in the eudiometer = 101.3 kPa
  • The partial pressure of H₂O = 2.41 kPa

The vapor pressure of H₂ gas = 101.3 kPa - 2.41 kPa

The vapor pressure of H₂ gas = (101.3 - 2.41) kPa

The vapor pressure of H₂ = 98.89 kPa

Therefore, we can conclude that the vapor pressure of H₂ is 98.89 kPa.

Learn more about partial pressure here:

brainly.com/question/14281129?referrer=searchResults

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A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of t
ioda

Answer:

0.053 L  is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

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