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Scorpion4ik [409]
2 years ago
14

What volume in milliliters of 0.0180 M Ba(OH)₂ is required to neutralize 55.0 mL of 0.0500 M HCl?

Chemistry
1 answer:
Nataly_w [17]2 years ago
5 0

The volume in milliliters of 0.0180 M Ba(OH)₂ that is required to neutralize 55.0 mL of 0.0500 M HCl is 152.78mL.

<h3>HOW TO CALCULATE VOLUME?</h3>

The volume of a solution can be calculated by using the following expression:

C1V1 = C2V2

Where;

  • C1 = initial concentration
  • C2 = final concentration
  • V1 = initial volume
  • V2 = final volume

The volume can be calculated as follows;

0.0180 × V1 = 55 × 0.05

0.0180V1 = 2.75

V1 = 2.75 ÷ 0.0180

V1 = 152.78mL

Therefore, the volume in milliliters of 0.0180 M Ba(OH)₂ that is required to neutralize 55.0 mL of 0.0500 M HCl is 152.78mL.

Learn more about volume at: brainly.com/question/1578538

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And in order to find the concentration we can use a figure called the "O2

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So then the final answer for this case would be an increase of 7%

Explanation:

For this case we know that a man with normal lungs have an arterial Po2 os 40 mm Hg.

Then we know that this man take an overdose of barbiturates thats halves his aveolar ventilation without changing his metabolism

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And we want to find hor much does his inspired oxyden concentration % have to increased to return his avelolar Po2 to the original level.

On this case we can apply a proportion rule and we have this:

\frac{4}{5} = \frac{40}{x}

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x = 40 * \frac{5}{4}= 50 mm Hg

So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.

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¡Hola!

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Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:

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