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Charra [1.4K]
1 year ago
8

In the laboratory, a student dilutes 18.9 ml of a 10.0 m perchloric acid solution to a total volume of 250.0 ml. what is the con

centration of the diluted solution?
Chemistry
1 answer:
Jobisdone [24]1 year ago
6 0

The concentration of the diluted solution is 0.756 M

Given,

Volume of stock solution (V1) = 18.9 mL

Molarity of stock solution (M1) = 10 M

Volume of diluted solution (V2) = 250 mL

M1V1 = M2V2

10*18.9=M2*250

M_{2}=0.756 M

<h3>Solution </h3>

A homogeneous mixture of two or more substances in relative proportions is referred to as a solution in chemistry. The limit of solubility is the point at which this mixture ceases to be a solution. Although the word "solution" is frequently used to refer to the liquid state of matter, solutions of gases and solids are also possible. Any of these three substances can be a solution. A mixture of liquids, gases, and solids can also make up a solution. Some solutions, like seawater, include a wide variety of solutes, including salts, oxygen, and organic compounds.

Learn more about solution here:

brainly.com/question/1616939

#SPJ4

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Give the nuclear symbol for the isotope of beryllium for which a=10? enter the nuclear symbol for the isotope (e.g., 42he).
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Read 2 more answers
When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
sp2606 [1]

Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

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