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Bezzdna [24]
3 years ago
14

A food web is shown below. Hawk Fox Snake Mouse Frog Grasshopper Beetle Fungus Grass Which organisms could be both predators and

prey depending on the relationship represented on the food web? A Hawk and fox B Grasshopper and beetle C Snake, frog, and mouse D Fungus and grass​

Physics
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

A

A fox and hawk can be both a predator and prey

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Technician A says that battery corrosion is commonly caused by spilled electrolyte. Technician B says that battery corrosion is
klio [65]

Answer:

B is correct

Explanation:

Battery contains sulfuric acid H2SO4. As the battery cools and eats with the running of the motor, hydrogen gasses is released from electrolytic reactions and escapes through the vents. it mixes with other particulates. The mixtures react at the electrical terminals of the battery and corrosion is formed.

Corrosion can be prevented by the addition of anti corrosion gel. It can also be removed by scrubbing the terminals with a hard brush

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4 years ago
What is the study of sound waves called?<br><br> short answer pls
sweet [91]

Answer:

Acoustics

Explanation: It’s the branch of physics that deals with the study of mechanical waves in gases, liquids, and solids including topics such as vibration, sound, ultrasound and infrasound.

7 0
3 years ago
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The North American and European plates of the
faust18 [17]
Distance = speed/time

width/distance given is 2.9*10^3 miles
1 mile=1.6*10^6mm
total distance now is 2.9*1.6*10^9
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5 0
3 years ago
What is the emf of this cell under standard conditions? Express your answer using three significant figures.
madreJ [45]

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is E^o_{cell}  = 2.35 V

Explanation:

From the question we are told that

   The ionic equation is  

               3 Ce^{4 +}  _{(aq)}  +  Cr _{(s)}  \to  3 Ce^{3+} _{(aq)}  +  Cr^{3r}  _{(aq)}

Now under standard conditions the reduction  half reaction  is

      Ce^{4+}  + e \to Ce^{3+}  ; \ \ E^o_r  =  1.61 V

And the oxidation half reaction is

      Cr^{3+}  +  3e^{-}  \to  Cr ; \    \  \  E^o_o  =  - 0.74 V

The emf of this cell under standard conditions  is mathematically represented as

     E^o_{cell}  =  E^o _r -  E^o _o

substituting values

     E^o_{cell}  =  1.61  - (- 0.74)

    E^o_{cell}  = 2.35 V

     

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3 years ago
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