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3 years ago

Conclusion: Summarize the results of this experiment in one or two sentences. What

Chemistry

1 answer:

GaryK [48]3 years ago

4 0

Answer:

The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic color the salt turns the flame of a Bunsen burner.

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when particles of matter are in direct contact and movement occurs in the substance due to convection, thermal energy transfers

liq [111]

Thermal energy transfers in a solid state, due to convection, in metalic substances.

This is because the covalent bonds between the atoms are being broken and reformed again while the metal is experiencing stress. <em>covalent bonds store energy. </em>

4 0

3 years ago

Which of the following statements best describes a difference between mechanical waves and electromagnetic waves?

kipiarov [429]

The correct answer is the 3rd one

6 0

3 years ago

A single serving bag of snack chips contains 65.0 Cal. Assuming that all of the energy from eating these chips goes toward keepi

AveGali [126]

Answer:

= 62.1 hours

Explanation:

Energy provide by the serving is 65 cal

= 65 cal × 4.184 Kj = 271.96 kJ

271.96 KJ = 271960 J

Energy required for 1minute of energy

= 73 x 1

= 73 J/min

So, 271960 joules will be required for 271960 heart beat

Minutes = 271960 / 73

= 3593.94 minutes

Time in hours = 3725.429 / 60

= 62.1 hours

5 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

Onsider the following reaction at equilibrium:

11111nata11111 [884]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$C(s)+H_2O(l)\leftrightharpoons CO(g)+H_2(g)$

A. C is added to the reaction mixture.

If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.

B. $H_2O$ is condensed and removed from the reaction mixture.

If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.

C. CO is added to the reaction mixture.

If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.

D. $H_2$ is removed from the reaction mixture.

If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.

5 0

3 years ago