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3 years ago

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

2 answers:

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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Calculate %ic of the interatomic bonds for the intermetallic compound al6mn. on the basis of this result, what type of interatom

pashok25 [27]

Answer : % ionic character is 0.20%

Bonding between the two metals will be purely metallic.

Explanation: For the calculation of % ionic character, we use the formula

$\% \text{ ionic character}= [1-e^{\frac{-(X_A-X_B)^2)}{4}}]\times(100\%)$

where $X_A$ & $X_B$ are the Pauling's electronegativities.

The table attached has the values of electronegativities, by taking the values of Al and Mn from there,

$X_{Al}=1.61$

$X_{Mn}=1.55$

Putting the values in the electronegativity formula, we get

$\% \text{ ionic character}= [1-e^{\frac{-(1.61-1.55)^2)}{4}}]\times(100\%)$

= 0.20%

Now, there are 3 types of inter atomic bonding

1) Ionic Bonding: It refers to the chemical bond in which there is complete transfer of valence electrons between atoms

2) Covalent Bonding: It refers to the chemical bond involving the sharing of electron pairs between 2 atoms.

3) Metallic Bonding: It refers to the chemical bond in which there is an electrostatic force between the positively charged metal ions and delocalised electrons.

In $Al_6Mn$ compound, the % ionic character is minimal that is $\sim$ 0.20% and there are two metal ions present, therefore this compound will have metallic bonding.

8 0

3 years ago

It is observed that the atoms of hydrogen in gas discharge tube emit radiations whose spectrum shows line characteristics (line

azamat

Answer: the line Spectra of hydrogen lies between the ultra-violet, visible light and infra-red of the electro magnetic spectrum

Explanation:

Electromagnetic radiation spans an wide range of wavelengths and frequencies. This range is called the electromagnetic spectrum. The electromagnetic spectrum is generally divided into seven regions, in order of decreasing wavelength and increasing energy and frequency. The 7 regions includes; radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.

lower-energy radiation, such as radio waves, is expressed as frequency while microwaves, infrared, visible and UV light are usually expressed as wavelength and finally, higher-energy radiation such as X-rays and gamma rays, is expressed in terms of energy per photon.

Therefore, hydrogen lies between the ultra-violet, visible light and infra-red region of the electro magnetic spectrum.

5 0

3 years ago

A covalent compound that shares two pairs of electrons.<br> What is it?

Nitella [24]

The two oxygen atoms share two pairs of electrons, so two covalent bonds hold the oxygen molecule together

3 0

3 years ago

What is the wavelength of light with 4.01 x 10^-19 J of energy? (The speed of light in a vacuum is 3.00 x 10^8 m/s, and Planck's

Vesna [10]

Answer:

hope it helps:)))!!!!!!

4 0

3 years ago

Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g)

GarryVolchara [31]

$\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \; \text{kJ} \cdot \text{mol}^{-1}$

Explanation

$\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)$

$\Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1}$
$\Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)
$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1}$ (from the question)

As an the most stable allotrope under standard conditions, $\Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1}$

By definition,

$\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products}) - \Delta H\textdegree{}_f (\text{all reactants})$

$\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \; \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\ = \Delta H\textdegree{}_{\text{rxn}}$

$\begin{array}{ccc} \Delta H\textdegree{}_f (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array}$

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the $\Delta H\textdegree{} _f$ given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

$\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \; \text{kJ} \cdot \text{mol}^{-1}$ (Cox, Wagman, et al., 1984)

... yields $\Delta H\textdegree{}_f (\text{IF} \; (g) ) \approx -64 \; \text{kJ}\cdot \text{mol}^{-1}$ , which deviates significantly from the experimental value of $-94.76 \; \text{kJ}\cdot \text{mol}^{-1}$ (Chase, 1998.) It is thus assumed that the $\Delta H\textdegree{}_\text{rxn}$ value provided requires a reaction with $\text{I}_2 \; (s)$ rather than $\text{I}_2 \; (g)$ as a reactant.

3 0

3 years ago