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Nata
1 month ago
6

A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is

the concentration of cu2+(aq ) in the solution?
Chemistry
1 answer:
castortr0y [3K]1 month ago
4 0
<span>Response: A 1.00 L solution that includes 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). Contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine. Using the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with double that amount of en = 0.000600 mol of en. Thus, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted leaves 0.00180 mol en unreacted. According to the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to yield an equivalent of 0.000300 moles of Cu(en)2^2+ The formation constant Kf for Cu(en)2^2+ is 1x10^20. Therefore, 1 Cu+2 and 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20 = [0.000300] / [Cu+2] [0.00180 ]^2 Solving for [Cu+2] gives [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Thus, Cu+2 = 9.26 e-19 Molar. Since Kf only has 1 significant figure, round that to 9 X 10^-19 Molar Cu+2.</span>
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Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K

Starting temperature of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

mC \Delta T = mC \times \Delta T

10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

T_{f} - 100^{o}C = 0^{o}C - T_{f}

2T_{f} = 100^{o}C

T_{f} = 50^{o}C

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

\Delta H = mC \Delta T

= 10000 g \times 0.385 J/K g \times 323 K

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Next, determine the entropy change for block 1 as follows.

\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

= 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

= 10000 g \times 0.385 J/K g \times -0.143

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

                   = -554.12 J/K + 647.49 J/K\Delta S_{total} = \Delta S_{1} + \Delta S_{2}

           = 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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castortr0y [3046]

Answer:

(C) The average speed of molecules in ethane is the same as that of propanol.

Explanation:

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Tems11 [2777]

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