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Gre4nikov [31]
3 years ago
12

Example of market structures of monopoly

Chemistry
1 answer:
IgorC [24]3 years ago
8 0

Answer: A monopoly is the absence of competition in the market.

Explanation:

In such circumstances, the market creates a monopoly of one producer who takes huge capital and dictates prices. An example of a monopoly on the market is the existence of only one company that makes up the entire economic branch. In such circumstances, the monopolist can increase the product's price without losing the entire sale, i.e., operating successfully. In that situation, the monopolist remains the only one on the market, and the competition has no access to the market.

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In a mixture of carbon dioxide in water ( a soft drink ) the Carbon Dioxide is the solute
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Answer:

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I need some help please
Norma-Jean [14]

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the first one to the third box

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What are the parts of a chromosome
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At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubil
sertanlavr [38]

Answer: Thus the solubility of N_2 gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of N_{2} in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,  C=k_{H}P.

k_{H}= Henry’s law constant = ?

The partial pressure (P) of N_{2} in water = 4.07 atm

\C= k_{H}\times P\\95.7mg=k_{H}\times 4.07

k_{H}=23.5

At pressure of 10.0 atm

C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg

Thus the solubility of N_2 gas in water, at the same temperature, is 235mg/100g

6 0
3 years ago
The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

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

3.0\times 10^{-23}=(0.220+s)\times (2s)^2

s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

7 0
3 years ago
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