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bija089 [108]
3 years ago
12

A chemist makes 620 mL of barium chloride working solution by adding distilled water to180mL of a 1.39mol/L stock solution of ba

rium chloride in water. Calculate the concentration of the chemist's working solution. Be sure your answer has the correct number of significant digits.
Chemistry
2 answers:
katen-ka-za [31]3 years ago
7 0

Answer:

The correct answer is 0.40 mol/L.

Explanation:

In order to solve the problem, we use the following expression, where V1 and C1 are the volume and concentration of stock solution and V2 and C2 are the volume and concentration of working solution:

V1 x C1= V2 x C2

⇒C2= V1 x C1/ v2

  C2= (180 ml x 1.39 mol/L)/ 620 ml

  C2= 0.40 mol/L

Svetllana [295]3 years ago
6 0

Answer:

0.404 mol / L

Explanation:

The <em>total moles of barium chloride</em> can be calculated from <em>the volume and concentration of the stock solution</em>:

1.39 mol/L * 0.180 L = 0.2502 moles barium chloride

Now we <u>calculate the concentration of the working solution, by dividing the total moles by the final volume</u>:

0.2502 mol / 0.620 L = 0.404 mol / L

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Answer:

see explanation...

Explanation:

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Protons (p⁺)                12                                 27                             17              

Neutrons (n⁰)             12                                 33                             18              

Electrons (eˉ)             10                                 24                             18              

                                  (c)                                 (b)                            (a)

                         12/2 : 12/2 : 10/2      27/3 : 33/3 : 24/3        #n⁰ = 18

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3 years ago
Why is most of the Earth’s fresh water unavailable?
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Answer:

2.5% of the earth's fresh water is unavailable: locked up in glaciers, polar ice caps, atmosphere, and soil; highly polluted; or lies too far under the earth's surface to be extracted at an affordable cost.

Explanation:

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Why is mercury the only metal to have been used in thermometers
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An IV bag is labeled as 0.800% w/w sodium chloride in water solution with a density of 1.036g/mL. What is the concentration of s
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Answer:

[NaCl] = 0.14 M

[NaCl] = 0.14 m

Mole fraction of NaCl → 2.48×10⁻³

Explanation:

This is a problem of concentration.

0.8% w/w means that in 100 g of solution, we have 0.8 g of solute, in this case NaCl.

Let's determine the volume with the density.

Solution density = Solution mass / Solution volume

1.036 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.036 g/mL → 96.5 mL

Let's calculate molarity (mol/L)

We convert the mass to moles (mass / molar mass)

0.8 g / 58.45 g/mol = 0.0137 moles

We must convert the volume of solution to L.

Molarity is mol/L, moles of solute in L of solution.

96.5 mL . 1L/1000 mL = 0.0965 L

0.0137 mol / 0.0965 L = 0.14 M

Let's determine molality and mole fraction.

Molality are moles of solute in 1 kg of solvent (mol/kg)

Mass of solution = Mass of solute + Mass of solvent

100 g = 0.8 g + Mass of solvent

100 g - 0.8 g = Mass of solvent → 99.2 g

Then, we must convert the mass of solvent to kg

99.2 g . 1kg / 1000 g = 0.0992 kg

Molality: 0.0137 mol / 0.0992 kg  → 0.14 m

Mole fraction → moles of solute / moles of solute + moles of solvent

Let's find out the moles of solvent ( mass / molar mass)

99.2g / 18 g/mol = 5.511 mol

Total moles = 5.511 + 0.0137 → 5.5247 moles

Mole fraction = 0.0137 / 5.5247 → 2.48×10⁻³

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