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Sholpan [36]
3 years ago
5

What is a liquid homogeneous mixture called in which the solute is dissolved and distributed evenly within the solvent? Colloid

Compound Emulsion Solution\
Chemistry
2 answers:
mestny [16]3 years ago
8 0
It would. be a Solution.


Tomtit [17]3 years ago
6 0

Hello The answer would be Solution I did the test you are on and made a 100 A+


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A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo
klio [65]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M

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Which bent shape has more repulsion and a smaller bond angle?<br> bent 6A<br> bent 5A
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Adding heat to water results in a relatively small temperature change because
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______________________________________________________
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3 years ago
A chemist must dilute 34.3mL of 1.72mM aqueous calcium sulfate solution until the concentration falls to 1.00mM. He'll do this b
EastWind [94]

Answer:

58.9mL

Explanation:

Given parameters:

Initial volume  = 34.3mL    = 0.0343dm³

Initial concentration  = 1.72mM   = 1.72 x 10⁻³moldm⁻³

Final concentration  = 1.00mM = 1 x 10⁻³ moldm⁻³

Unknown:

Final volume  =?

Solution:

Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.

   Therefore;

              C₁V₁  =  C₂V₂

where C and V are concentration and 1 and 2 are initial and final states.

        now input the variables;

                      1.72 x 10⁻³ x  0.0343 = 1 x 10⁻³  x V₂

                        V₂ = 0.0589dm³ = 58.9mL

         

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