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GaryK [48]
2 years 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:
Alik [6]2 years ago
4 0
<span>Answer: A 1.00 L solution containing 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 by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
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Hello!

In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy:

s_1=s_{f,175kPa}+q*s_{fg,175kPa}\\\\s_1=1.4850\frac{kJ}{kg*K} +0.1*5.6865\frac{kJ}{kg*K}=2.0537\frac{kJ}{kg*K}

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Next step is to compute the mass of steam given the specific volume of steam at 175 kPa and the 10% quality:

m_1=\frac{0.028m^3}{(0.001057+0.1*1.002643)\frac{m^3}{kg} } =0.274kg\\\\m_2=\frac{0.028m^3}{(0.001053+0.4*1.158347)\frac{m^3}{kg} } =0.0603kg

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Whereas sfg stands for the entropy of the leaving steam to hold the pressure at 150 kPa and must be greater than 0; thus we plug in:

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