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lara31 [8.8K]
3 years ago
12

A zinc coating is added to the surface of a tin component to provide corrosion protection to the tin. Assume that the coating ad

heres to (physically contacts) the underlying tin. A small region of the zinc coating is chipped, but the remainder of the coating remains intact. What do you expect will happen to the coating and the exposed tin in the presence of salt water and oxygen?
Chemistry
1 answer:
Norma-Jean [14]3 years ago
5 0

Answer:

In the presence of salt water and oxygen the coating will not corrode. As long as zinc coating is present and remains intact corrosion is not possible.

Explanation:

Here it is given that a tin is present so firstly tin is made of a chemical element

which belongs to carbon family in periodic table of group 14.

It is a silvery,soft, white metal with a bluish tinge.

Now the covering which is been done on the tin is Zinc.

so, zinc is known to be served as a sacrificial coater.

Their is an amazing reason behind zinc coating being so effective and intact  i.e. Its own corrosive properties are such that it stops corrosion.

Their is a process which is known as a galvanic corrosion which refers to that "ZINC" defers to the metal  to which it is protecting.

It is even more electrochemically active than iron itself.

Here, it is mentioned that zinc coating gets chipped but the coating remains intact. So, if the zinc is not removed from the tin's surface it will not get corroded when it is exposed to salt water and oxygen.

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What volume will 2.5 mol of a gas at STP occupy ?
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56

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Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2
lapo4ka [179]

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Explanation:

Moles of  CO and Cl_2 = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of CO and Cl_2  =\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M

The given balanced equilibrium reaction is,

                            CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.          0.860M     0.860M           0

At eqm. conc.    (0.860-x) M  (0.860-x) M     (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

Now put all the given values in this expression, we get :

4.95=\frac{x}{(0.860-x)^2}

By solving the term 'x', we get :

x =  0.532 M

Thus, the concentrations of CO,Cl_2\text{ and }COCl_2 at equilibrium are :

Concentration of CO = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of Cl_2 =  (0.860-x) M  =  (0.860-0.532) M = 0.328 M

Concentration of COCl_2 = x M = 0.532 M

3 0
3 years ago
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