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oksano4ka [1.4K]
3 years ago
11

Why do you think the process to create a protected area takes so long?

Chemistry
1 answer:
Paladinen [302]3 years ago
4 0

The process to create a protected area takes a long time because there's numerous things that have to be taken in account and sorted out before the area gets that kind of legal status.

Usually there's people living in the area that is supposed to be protected, though it is mostly small settlements. The people living there are connected with that area, use the resources, and their lives are largely based and dependent on it. In order for any problems to be avoided, these people should all be educated, deals to be made with them, and properly negotiated so that the area wouldn't have any damage, but also them to be able to sustain themselves.

The boundaries of the area are also complicated, it it takes time to sort out what parts should be included and what not. Basically to set the priorities.

People should be employed to protect and take care of the area, so they have to undergo a training, education, and to get familiar with it.

The legalization of the area as such can sometimes too make problems and take some time.

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When aqueous solutions of magnesium nitrate and sodium carbonate are mixed together what will happen
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When sodium carbonate and magnesium nitrate solutions are combined, a precipitate is created as a result of a double-displacement process. As you balance the chemical equation, make predictions about the reaction's byproducts.

<h3>What occurs when magnesium sulfate and sodium carbonate are combined?</h3>

A solid may form when the solutions of two soluble salts are combined. Salts that are insoluble are created during precipitation processes. When magnesium sulfate and sodium carbonate are combined, magnesium carbonate, an insoluble salt, results.

<h3>What happens when hydrochloric acid and sodium carbonate interact?</h3>

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to know more about magnesium and sodium carbonate here:

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4 0
1 year ago
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3 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


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