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RSB [31]
3 years ago
12

What is the molarity of a sodium hydroxide solution containing 2 moles of NaOH with a volume of .5L?

Chemistry
1 answer:
Maksim231197 [3]3 years ago
5 0

Answer:

0.4M NaOH

Explanation:

Molarity, M, is an unit of concentration widely used defined as the ratio between moles of solute (In this case, NaOH) and volume of solution in liters.

As the solution contains 2 moles of NaOH-Moles of solute- in 5L of solution, the molarity is:

2 moles NaOH /  5L =

<h3>0.4M NaOH</h3>
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A stock solution has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L. What was the volu
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The volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.

<h3>How to calculate volume?</h3>

The concentration of a solution can be calculated using the following formula:

C1V1 = C2V2

Where;

  • C1 = initial concentration = 1.5M
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  • V1 = initial volume = ?
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1.5 × V1 = 0.54 × 0.18

1.5V1 = 0.0972

V1 = 0.0972 ÷ 1.5

V1 = 0.065L

Therefore, the volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.

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

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2 years ago
A scientist spilled a few drops of dilute hydrochloric acid (hcl) on a lab table. for safety purposes, the scientist sprinkled s
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Hello!

The observation that would provide the best evidence that a chemical reaction occurred is that The baking soda and hydrochloric acid combined, and bubbles formed.

When baking soda (NaHCO₃) and Hydrochloric Acid (HCl) combine, the following reaction happens:

NaHCO₃ + HCl → NaCl + H₂O + CO₂(g)↑

The gaseous Carbon Dioxide (CO₂) generated in this reaction is the responsible for the bubbles. The releasing of this gas is an evidence that a chemical reaction occurred between NaHCO₃ and HCl.

Have a nice day!
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This is a material that allows heat/electricity to transfer.
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A material that allows heat/electricity to transfer is called a conductor.
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