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Radda [10]
3 years ago
11

What is the molarity of a solution containing 3.6 mol of KCl in 750 mL?

Chemistry
1 answer:
Vesnalui [34]3 years ago
4 0

Answer:

\boxed {\boxed {\sf 4.8 \ M \ KCl}}

Explanation:

The molarity of a solution is found by dividing the moles of solute by the liters. of solvent.

molarity=\frac{moles}{liters}

We know the solution has 3.6 moles of potassium chlorine. We know there are 750 milliliters of solvent, but we need to convert this to liters.

  • 1 liter is equal to 1000 milliliters. Set up a proportion.
  • \frac {1 \ L }{1000 \ mL}
  • Multiply by 750 mL and the units of mL will cancel.
  • 750 \ mL * \frac {1 \ L }{1000 \ mL} = \frac {750 \ mL}{1000 \ mL}= 0.75 \ L

Now we know the moles and liters, so we can calculate molarity.

  • moles= 3.6 mol KCl
  • liters= 0.75 L

Substitute these values into the formula.

molarity= \frac{3.6 \ mol \ KCl}{0.75 \ L}

Divide.

molarity= 4.8 \ mol \ KCl/ L

  • 1 mole per liter is equal to 1 molar.
  • Our answer of 4.8 moles of potassium chloride per liter is equal to 4.8 M KCl

molarity= 4.8 \ M \ KCl

The molarity of the solution is <u>4.8 M KCl.</u>

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Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

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-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

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-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

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The reaction quotient for this reaction is as follows.

            Q = \frac{[Fe^{2+}]^{2}[Zn^{2+}]}{[Fe^{3+}]^{2}}

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