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leva [86]
3 years ago
5

What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?

Chemistry
1 answer:
Neporo4naja [7]3 years ago
8 0

Answer:

168.75 ml

Explanation:

M1V1=M2V2

375ml*.45M=1M*V2

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lora16 [44]

Answer:

0.77093576

Explanation:

4 0
3 years ago
What is the approximate depth of the calcite compensation depth (CCD) in the ocean? View Available Hint(s) What is the approxima
masya89 [10]

Answer:

The correct answer to the question is

D. 3 miles (4.5 kilometers)

Explanation:

The Carbonate compensation depth or (CCD) is the ocean depth at which calcite, (calcium carbonate) dissolves. At the CCD, the solvation rate of calcite is greater than the supply rate, such that all calcite are consumed.

The carbonate compensation depth varies in different parts of the ocean and can be reached at about 3 miles or 4.5 Kilometers.

4 0
3 years ago
What's a material that contains oxygen and silicon??
Maslowich
Sand/quartz or silicon dioxide contains both silicone and oxygen

hope that helps
8 0
2 years ago
Identify the oxidizing and reducing agent in the following reaction, and determine which element is oxidized and which is reduce
erma4kov [3.2K]

Answer:

Explanation:

Fe⁺²(aq) + ClO₂(aq) → Fe⁺³(aq) + ClO₂⁻(aq)

Here oxidation number of Fe is increased from +2 to +3 , so Fe is oxidised .

The oxidation number of Cl is reduced from + 4 to +3  so Cl is reduced .

So ClO₂(aq) is oxidising agent and Fe⁺²(aq) is reducing agent .

8 0
3 years ago
You need to determine the mass of an aqueous solution. You determine the mass of your 10.0 mL graduated cylinder to be 23.731 g.
Amanda [17]

Answer: 0.86g/mL

Explanation:

Mass of empty cylinder = 23.731g

Mass of cylinder + liquid = 26.414g

Mass of the liquid = 26.414 — 23.731

= 2.683g

Volume of the liquid = 3.12mL

Density = Mass / volume

Density = 2.683g / 3.12mL

Density = 0.86g/mL

3 0
3 years ago
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