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Maurinko [17]
3 years ago
13

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

alloy that contains 25 wt% Au and 75 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively. The atomic weight of Au is 196.97 g/mol.

Engineering
1 answer:
evablogger [386]3 years ago
5 0

Answer:

Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3

Explanation:

The step by step and appropriate substitution is as shown in the attachment.

From number of moles = Concentration x volume

number of moles = number of particles/ Avogadro's number

Volume = mass/density, the appropriate derivation to get the number of moles of atoms

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At user equilibrium, $t_2=t_1$

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Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

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150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

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divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

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