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Ivahew [28]
3 years ago
6

For an alloy that consists of 78.2 g copper, 103.5 g zinc, and 2.8 g lead, what are the concentrations of (a) Cu, (b) Zn, and (c

) Pb in weight percent? The atomic weights of Cu, Zn, and Pb are 63.54, 65.39, and 207.2 g/mol, respectively.
Chemistry
1 answer:
TEA [102]3 years ago
8 0

Answer:

Explanation:

Mass percent is defined as the mass of an element divided by the sum of masses of all the elements multiplied by 100. It is generally used to define the concentration. It does not depend on concentration.

It is given as;

Mass percent = (mass of an element / Total mass of the compound ) × 100

Mass of  compound (alloy) = 78.2 g copper  + 103.5 g zinc +2.8 g lead = 184.5 g

(a) Cu

Mass of Cu = 78.2

Mass percent = (78.2 / 184.5) * 100 = 42.38%

(a) Zn

Mass of Zn = 103.5

Mass percent = (103.5 / 184.5) * 100 = 56.10%

(c) Pb

Mass of Pb = 2.8

Mass percent = (2.8 / 184.5) * 100 = 1.52%

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At 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

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G = H + TS

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3) conditions

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ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

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ΔG = 0 =>  T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at  T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

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                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

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where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

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Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

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