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AnnZ [28]
3 years ago
15

A standard solution contained 0.8 mg/mL. A student took 2 mL of the standard solution and added 10 mL of water. What is the new/

final concentration?
Chemistry
2 answers:
galben [10]3 years ago
7 0
To get the concentration of the second solution let us use the following formulae

C1V1=C2V2 where C1 is concentration of first solution and V1 is the volume of solution first solution. on the other hand C2 is the concentration of second solution and V2 is the volume of second solution.

therefore

0.8×2=(2+10)×C2
   1.6 =12×C2
1.6/12=C2
C2     = 0.1333mg/mL
nikdorinn [45]3 years ago
4 0
<span>If the standard solution contains 0.8 mg/mL of a substance and there are 2 mL of the solution, then that quantity contains 0.8*2 = 1.6 mg of the substance. Adding 10mL of water, the new volume will be 2mL + 10mL = 12 mL and there will still be the same 1.6 mg of the substance. Thus the new concentration will be 1.6mg/12mL = 0.133 mg/mL.</span>
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lyudmila [28]

Zn, Cd, and Ag are transition metals that usually form only one monoatomic cation.

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These cations form only one type of ion, while iron and copper form more than one type of cations.

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5 0
1 year ago
Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
madreJ [45]

Answer:

Molar \ solubility=3.12x10^{-5}M

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-

The equilibrium expression is:

Ksp=[Ca^{2+}][F^-]^2

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent x is computed as follows:

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Thus, the molar solubility equals the reaction extent x, therefore:

Molar \ solubility=3.12x10^{-5}M

Regards.

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3 years ago
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