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ohaa [14]
3 years ago
12

Please help me for question 1 and 2

Chemistry
1 answer:
likoan [24]3 years ago
4 0

Answer:-

1) 6 mol

2) Mo

Explanation: -

Mass of Ozone = 48 g

Chemical formula of ozone = O3

Molar mass of Ozone O 3 = 16 x 3 = 48 g mol-1

Number of moles of ozone = Mass / molar mass

= 48 g / 48 g mol-1

= 1 mol

According to Avogadro’s law, 1 mole of a substance has 6.02 x 10^ 22 molecules.

So 1 mol of O3 has 6.02 x 10^ 22 molecules of ozone.

Now each Ozone molecule has 3 atoms of oxygen.

So, 1 mol of ozone has 3 x 6.02 x 10^22 atoms of oxygen.

Sodium must have 2 x 3 x 6.02 x 10^22 atoms as per the question.

According to Avogadro’s law, 6.02 x 10^ 22 atoms are in 1 mol of sodium

So, for 2 x 3 x 6.02 x 10^22 atoms, there should be (1/ 6.02 x 10^ 22) x 2 x 3 x 6.02 x 10^22

= 6 mol of sodium.

b)

Let the mass of M be m g

Formula of hexafluoride = MF6.

Mass of the hexafluoride = g + 6 x 19

= m + 114

Mass of M=0.250g

Moles of M = 0.250/m

Mass of MF6= 0.547g

Moles of MF6 = 0.547/ (m + 114)

We know 1 mole of M gives 1 mole of MF6.

0.250/m moles of M gives 0.250/m moles of MF6.

But number of moles of MF6 = 0.547/ (m + 114)

Thus

0.250/m = (0.547)/ (m +114))

0.250m + 0.250 x 114 = 0.547m

m = 0.250 x 114 / (0.547 -0.250)

= 96

We see from the given data that Mo is 96.

So M is Mo.

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Assume that a firm has the following marginal benefit of pollution (denoted E for emissions, measured in tons): MB=150-5 E e. No
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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

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E=60-8MB/15

MB=112.5-1.875E

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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.

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