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inna [77]
3 years ago
10

Which compound plays an important role in leather processing and electroplating?

Chemistry
2 answers:
Margaret [11]3 years ago
7 0

<u>Answer;</u>

  • Hydrochloric acid

<u>Explanation;</u>

  • Hydrochlorc acid plays a vital role in leather processing and electroplating.
  • Hydrochloric acid has a number of important applications in real life these includes; fertilizers, and dyes, textile, in electroplating, rubber and in the photographic industries.
  • Electroplating is the process that involves coating a metal using another metal for the purpose of improving their appearance or preventing rusting.
  • Hydrochloric acids is used in the processing of leather or the tanning of leather process, in which tanneries use the acid to stop the growth of bacteria and also maintain proper pH.

abruzzese [7]3 years ago
5 0

Answer:

hydrochloric acid is the correct answer.

Explanation:

hydrochloric acid compound plays an important role in leather processing and electroplating.

Electroplating is a process in which metal is coated with a thin layer of another metal.hydrochoric acid is used in electroplating as it enables oxidation of copper and removes the copper ions from the solution.

Hydrochloric acid is used in leather processing because tanneries use hydrochloric acid to stop the growth of bacteria and to control the pH level of the leather.

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Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
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The unit cell edge length for the alloy is 0.405 nm

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concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

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density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
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