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jenyasd209 [6]
3 years ago
7

All of the following statements about the nitrogen cycle are incorrect, EXCEPT which?

Chemistry
1 answer:
Lapatulllka [165]3 years ago
8 0
Dentrifictation occurs when there is a short fall of available nitrogen in the soil
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When 0.250 moles of KCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change is observed. Given
777dan777 [17]

Explanation:

Upon dissolution of KCl heat is generated and temperature of the solution raises.

Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

             17.24 kJ/mol \times 0.25 mol

                = 4.31 kJ

or,             = 4310 J      (as 1 kJ = 1000 J)

Mass of solution will be the sum of mass of water and mass of KCl.

       Mass of Solution = mass of water + (no. of moles of KCl × molar mass)

                                    = 200 g + (0.25 mol \times 54.5 g/mol)

                                    = 200 g + 13.625 g

                                    = 213.625 g

Relation between heat, mass and change in temperature is as follows.

                             Q = mC \Delta T

where,    C = specific heat of water = 4.184 J/g^{o}C

Therefore, putting the given values into the above formula as follows.

                     Q = mC \Delta T

            4310 J = 213.625 g \times 4.184 J/g^{o}C \times \Delta T      

              \Delta T = 4.82^{o}C

Thus, we can conclude that rise in temperature will be 4.82^{o}C.

6 0
3 years ago
Assume that a firm has the following marginal benefit of pollution (denoted E for emissions, measured in tons): MB=150-5 E e. No
jeka57 [31]

Answer:

Explanation:

E)cost of pollution is reduction in benefit of the firms.

MB=150-5E,. MB=90-3E

E=30-MB/5. E=30-MB/3

Joint MB,

E=60-8MB/15

MB=112.5-1.875E

Total pollution reduction=24

Total pollution=60-24=36

MB=112.5-1.875*36=112.5-67.5=45

Firm 1

MB=150-5E.

45=150-5E.

E=-105/-5=21

Reduction=30-21=9

Firm 2,

MB=90-3E

45=90-3E

E=-45/-3=15

Reduction =30-15=15

So firm 2 is reducing 15 units of pollution and firm 1 reducing 9 units of pollution.

As each firm have to reduce 12 units of pollution but firm 2 reducing 3 units more while firm 1 reducing 3 less units.

So, firm 2 is selling 3 units of pollution emission permit to firm 1.

F)firm 1 reduce 9 units and firm 2 reduce 12 units of pollution after trade.

Total cost of pollution reduction will total Benefit reduction by pollution reduction.

MB=112.5 -1.875E

TB=112.5E-0.9375E^2

TB at E=60

TB=112.5*60-0.9375*60*60=6750-3375=3375

TB at E=36

TB=112.5*36-0.9375*36*36=4050-1215=2836

Total cost of pollution reduction=3375-2836=540

G)price of permit= cost of extra pollution reduction by firm 2 or total cost from 9 to 12 by firm 2

MB=90-3E

TB=90E- 1.5E^2

TB at E=18

TB=90*18 -1.5*18*18=1620-486=1134

TB at E=15

TB=90*15 -1.5*15*15=1350-337.5=1012.5

Permit price=1134-1012.5=121.5

Total cost to firm 2 =1134

Net total cost to firm 2=1134-121.5=1012.5

Total cost to firm 1=150E-2.5E^2=150*9-2.5*9*9=1350-202.5=1147.5

Net total cost=1147.5+121.5=1269

H) the total cost is lower in cap & trade policy is because the firm who has higher cost of pollution reduction is paying the other firm who has lower cost of pollution reduction to reduce more pollution ,so that his part of pollution reduction can be completed.

And the amount he is paying is equal to the amount that is cost of other firm of reducing additional pollution units.

So the cost the firm is lower as he is paying lower amount than if he reduce pollution by itself.

5 0
3 years ago
Phosphorus-32 is radioactive and has a half life of 14.3 days. What percentage of a sample would be left after 12.6 days
nadya68 [22]

Answer:

There is 54.29 % sample left after 12.6 days

Explanation:

Step 1: Data given

Half life time = 14.3 days

Time left = 12.6 days

Suppose the original amount is 100.00 grams

Step 2: Calculate the percentage left

X = 100 / 2^n

⇒ with X = The amount of sample after 12.6 days

⇒ with n = (time passed / half-life time) = (12.6/14.3)

X = 100 / 2^(12.6/14.3)

X = 54.29

There is 54.29 % sample left after 12.6 days

8 0
3 years ago
How is dry ice different from ordinary ice?
True [87]

Answer: Well "ordinary ice" is made from water (as we all know) and dried ice is made from carbon dioxide gas

Explanation:

•Hope this helps•

5 0
3 years ago
What is the maximum amount in moles of P2O5 that can theoretically be made from 235 g of P4 and excess oxygen?
Lady_Fox [76]
The correct answer would be 3.8mol
5 0
3 years ago
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