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Vladimir [108]
1 year ago
9

The concentration of an HCl acid solution is initially 6mol/L. What would the concentration of a new solution be if 200 mL of th

e original HCl solution is diluted with 1L of water.
Chemistry
1 answer:
pochemuha1 year ago
5 0

Taking into account the definition of dilution, the concentration of the new solution is 1 mol/L.

<h3>Dilution</h3>

When it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

<h3>Final volume</h3>

In this case, you know:

  • Ci= 6 mol/L
  • Vi= 200 mL
  • Cf= ?
  • Vf= 1 L (1000 mL) water + 200 mL of HCL= 1200 mL

Replacing in the definition of dilution:

6 mol/L× 200 mL= Cf× 1200 mL

Solving:

(6 mol/L× 200 mL)÷ 1200 mL= Cf

<u><em>1 mol/L= Cf</em></u>

In summary, the concentration of the new solution is 1 mol/L.

Learn more about dilution:

brainly.com/question/6692004

brainly.com/question/16343005

brainly.com/question/24709069

#SPJ1

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Answer : Broadly solids are divided into three categories;

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ii) Amorphous solids have a random structure, with little unorganized pattern long-range order.

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8 0
3 years ago
Read 2 more answers
What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?
Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})

Hence, the probability of finding the particle in the  one-dimensional box is:

P = \int_{x_{1}}^{x_{2}} \psi^{2} dx

P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx

P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]

P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})    

Solving for n=4:

P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})    

P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})    

P = \frac{1}{8}

I hope it helps you!

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Answer:

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Kisachek [45]

Answer:

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Hello!  These are the correct answers! (I took the K12 quiz)

Have a blessed day! :)

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