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Len [333]
3 years ago
8

Be sure to answer all parts. find the molar solubility of bacro4 (ksp= 2.1 × 10−10) in (a) pure water × 10 m (b) 1.6 × 10−3 m na

2cro4 × 10 m
Chemistry
1 answer:
Vlad [161]3 years ago
3 0
A) in pure water :

by using ICE table:

According to the reaction equation:

            BaCrO4(s)    →  Ba^2+(aq)    +   CrO4^2-(aq)

initial                               0                          0

change                          +X                       +X 

Equ                                  X                         X


when Ksp = [Ba^2+][CrO4^2-]

by substitution:

2.1 x 10^-10 = X* X

∴X = √2.1 x 10*-10

∴X = 1.4 x 10^-5

∴ the solubility = X = 1.4 X 10^-5

B) In 1.6 x 10^-3 m Na2CrO4

 by using ICE table:

According to the reaction equation:

            BaCrO4(s)  →  Ba^2+(aq)    +   CrO4^2-(aq)

initial                                 0                      0.0016

Change                           +X                      +X

Equ                                   X                      X+0.0016

when Ksp = [Ba^2+][CrO4^2-]

by substitution:

2.1 x 10^-10 = X*(X+0.0016) by solving for X 

∴ X = 1.3 x 10^-7

∴ solubility =X = 1.3 x 10^-7

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4 0
3 years ago
A 0.885 M solution of KBr whose initial volume is 82.5 mL has more water added until its concentration is 0.500 M. What is the n
4vir4ik [10]

Answer:

V_2=146mL

Explanation:

Hello there!

In this case, since the equation for the calculation of dilutions is:

M_1V_1=M_2V_2

Whereas M is the molarity and V the volume, because the final concentration is lower than the initial. Thus, since we are asked to calculate the final volume, we solve for V2 as follows:

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Answer:

d) osmotic pressure equal to that of blood

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hichkok12 [17]
I think the answer is c. two molecules
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