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Anna71 [15]
2 years ago
11

A saturated solution of pbcl2 is found to contain 2.88 x 10-2 m chloride ions. what is the correct value of the solubility produ

ct, ksp
Chemistry
1 answer:
bazaltina [42]2 years ago
4 0

The solubility equilibrium of PbCl_{2}:

PbCl_{2}(aq)  Pb^{2+}(aq) + 2 Cl^{-}(aq)

K_{sp}=[Pb^{2+}][Cl^{-}]^{2}

[Cl^{-}] = 2.88 * 10^{-2} M

[Pb^{2+}]=\frac{[Cl^{-}]}{2} = \frac{2.88 * 10^{-2}}{2}=1.44 *10^{-2}

K_{sp}=[Pb^{2+}][Cl^{-}]^{2}

= (1.44 * 10^{-2})(2.88*10^{-2})^{2}

= 1.19 * 10^{-5}

So, the corrected solubility product will be 1.19 * 10^{-5}

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Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ
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To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
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7 0
3 years ago
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3 0
3 years ago
What is the mass of 2.2x10^9 molecules of CO2? *​
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3 years ago
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