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3 years ago

What minimum concentration of clâ is required to begin to precipitate pbcl2? for pbcl2, ksp=1.17Ã10â5?

1 answer:

5 0

Missing in your question : [Pb2+] = 0.085 M

So from this balanced equation:
PbCl2 ↔ Pb+2 + 2Cl-

when Ksp = [Pb+2][Cl]^2
when we have Ksp = 1.17x10^-5 & [Pb+2] = 0.085 M so by substitution:
1.17x10^-5 = 0.085 [Cl-]^2

∴[Cl]^2 = (1.17x10^-5) / 0.085 = 1.38x10^-4
∴[Cl-] = √0.0002
= 0.0117 M
So 0.0117 M is the minimum concentration requires precipitating PbCl2

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Construct a three-step synthesis of 1,2-epoxycyclopentane from cyclopentanol by dragging the appropriate formulas into the bins.

zubka84 [21]

Answer:

(1) Bromination, (2) E2 elimination and (3) epoxidation

Explanation:

In the first step, -OH group in cyclopentanol is replaced by more facile leaving group Br by treating cyclopentanol with $PBr_{3}$
In the second step, E2 elimination in presence of strong base e.g. NaOEt/EtOH produce cyclopentene
In the third step, treatment of cyclopentene with mCPBA produces 1,2-epoxycyclopentane
Full reaction scheme has been shown below

3 0

3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

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3 years ago

Can artificial selection aid in the evolution of an organism?

stepan [7]

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3 years ago

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Answer:

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Alexeev081 [22]

Answer:

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3 years ago