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erma4kov [3.2K]
3 years ago
5

2) What is the concentration of NaCl in a solution prepared by diluting 56.98 ml of 0.5894 M stock

Chemistry
1 answer:
RSB [31]3 years ago
3 0

Answer:

The NaCl concentration will be 0.03 M.

Explanation:

Given data:

Initial volume = V₁ = 56.98 mL (56.98/1000 = 0.05698 L)

Initial concentration = M₁= 0.5894 M

Final volume = V₂= 1.20 L

Final concentration = M₂= ?

Solution:

By diluting the solution volume of solution will increase while number of moles of solute remain the same.

Formula:

Initial concentration × Initial volume  = Final concentration × Final volume

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 0.5894 M × 0.05698 L / 1.20 L

M₂ = 0.0336 M /1.20  

M₂ = 0.03 M

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If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit
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<u>Answer:</u>

<u>For a:</u> The volume of the box is 217.5 mL

<u>For b:</u> The volume of the box is 0.2175 L

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To calculate the volume of cuboid, we use the equation:

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To convert the volume of cuboid into milliliters, we use the conversion factor:

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