How do the isotopes of an element differ?

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

How many moles of oxygen are required to produce 37.15 g CO2? 37.15 g CO2 = mol O2

NNADVOKAT [17]

Answer:

0.84 moles of oxygen are required.

Explanation:

Given data:

Mass of CO₂ produced = 37.15 g

Number of moles of oxygen = ?

Solution:

Chemical equation:

C + O₂ → CO₂

Number of moles of CO₂:

Number of moles = mass/molar mass

Number of moles = 37.15 g/ 44 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of oxygen and carbon dioxide.

CO₂ : O₂

1 : 1

0.84 : 0.84

0.84 moles of oxygen are required.

6 0

3 years ago

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To determine the age of fairly recent fossils and organic artifacts, it is possible to analyze the amounts of the isotopes 14C a

dolphi86 [110]

Answer: The net change in the atoms is the conversion of a neutron to a proton, turning Carbon (6 protons) into Nitrogen (7 protons).

Explanation:

Carbon-14, generated from the atmosphere, has 6 protons and 8 neutrons. That's where the 14 comes from, called the mass number, is the sum of protons and neutrons (6+8=14).

Carbon-14 is radioactive and decays by beta decay. That means one of its neutrons spontaneously turns into a proton, an electron, and a neutrino, according to:

$^{14} C \rightarrow\ ^{14} N\ +\ e^- +\ neutrino$

After that, the atom has 7 protons and 7 neutrons, maintaining its mass number but changing its atomic number from 6 to 7, turning into Nitrogen.

3 0

3 years ago

An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i

lutik1710 [3]

Answer:

$C_{metal}=126.6\frac{J}{g\°C}$

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

$Q_{metal}=-Q_{water}$

Which can be also written as:

$m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})$

Thus, since we need the specific heat of the metal, we solve for it as shown below:

$C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}$

Best regards.

7 0

2 years ago

Match each term to its description. (3 points)

madreJ [45]

Answer: Limiting reactant = 3

Theoretical Yield= 1

Excess reactant=2

Explanation: The theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction. It can be calculated from: the balanced chemical equation. the mass and relative formula mass of the limiting reactant , and. the relative formula mass of the product.

An excess reactant is a reactant present in an amount in excess of that required to combine with all of the limiting reactant. It follows that an excess reactant is one remaining in the reaction mixture once all the limiting reactant is consumed.

The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated

3 0

3 years ago

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