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Studentka2010 [4]
3 years ago
6

Which of the following best explains why electroplating is a useful process in many industries?

Chemistry
2 answers:
UkoKoshka [18]3 years ago
8 0
Since I cannot find the choices, I will tell you some of the benefits of electroplating in different industries. You can compare these benefits with the choices you have and choose the best fit.

1- Forms a protective layer to protect the material from the conditions of the atmosphere as the corrosion

2- Improves the appearance of some inexpensive materials and makes them look more appealing

3- Can enhance the electrical conductivity of materials

4- electroplating of zinc-nickel or gold can survive high temperatures

5- Sometimes hardens the material

6- Increases the thickness of the material 
lapo4ka [179]3 years ago
3 0

Answer: A.

Explanation:  it makes some inexpensive materials look more appealing

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What are dipoles, and what is the difference between a natural dipole and an induced dipole?
Tasya [4]

Answer:

See explanation below.

Explanation:

Dipoles are molecules that have partial charges. It happens because of the difference in electronegativity of the elements. This property is the tendency that the atom has to take the electron to it, so, in the covalent bond, the shared pair of electrons is easily found at the more electronegativity atom, and so, it has a partial negative charge, and the other, a partial positive charge. This is a natural dipole.

If the difference of electronegativity is 0, or extremely close to 0, then the molecule is nonpolar, and so the molecule doesn't have partial charges. But, to be joined together and form the substance, the partial charge must be induced, so it's an induced dipole.

7 0
3 years ago
Help me please i'm begging you its DUE today its super easy
NARA [144]

Answer:

The answer will be listed below.

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3 years ago
Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

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3 years ago
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3 years ago
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aleksandrvk [35]

Answer:

24.07

Explanation:

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3 years ago
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