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USPshnik [31]
4 years ago
15

What is the molar solubility of magnesium fluoride in a solution that is 0.40 m f− ions? the ksp of mgf2 is 6.4 × 10−9. hints wh

at is the molar solubility of magnesium fluoride in a solution that is ions? the of is . 8.0 × 10−9 8.0 × 10−5 1.6 × 10−8 4.0 × 10−8?
Mathematics
1 answer:
velikii [3]4 years ago
6 0
<span>Answer: MgF2 ====⇒ Mg+2 + 2F- Ksp = [Mg+2][F-]^2 Let S = molar solubility of MgF2 [Mg+2] = S ; [F-] = 2s Since the [F-] is initially 0.40 M, then [F-] = 0.40 + 2S 6.4 x 10^-9 = (S) (0.40 + 2S)^2 ; one can neglect the 2S in the 0.40 + 2S expression since it is very, very small compared to the 0.40 already present. 6.4 x 10^-9 = S(0.40)^2 S = 4.0 x 10^-8 Molar solubility = 4.0 x 10^-8</span>
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postnew [5]

2  \frac{5}{8}  \div 2  \frac{1}{2}
Change the mixed fractions to improper fractions.
To do this, multiply the whole number part by the denominator. Add that to the numerator. Then write the result on top of the denominator.

You would now have:
\frac{21}{8}  \div  \frac{5}{2}
Change the division sign to a multiplication sign by turning the second fraction upside down.
That is,
\frac{21}{8}  \times  \frac{2}{5}

The answer is:
\frac{42}{40}

This can be simplified to:
\frac{21}{20}



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