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kenny6666 [7]
3 years ago
13

Photosynthesis and cellular respiration have a unique relationship in that the products of one process are the reactants of the

other. For example the products of photosynthesis are reactants of cellular respiration.
Select ALL of the molecules that are produced in photosynthesis and then used in cellular respiration.

A) ATP

B) water

C) oxygen

D) glucose

E)carbon dioxide​
Chemistry
2 answers:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

the answer is c) oxygen and D) glucose. Read the question correctly. it says what is *produced* in photosynthesis and *used* in cellular respiration.

Explanation:

photosynthesis produced oxygen and glucose to which cellular respiration uses (glucose and oxygen) to produce carbon dioxide and water. so carbon dioxide and water ARE NOT the answer to this.

also I took the test so yeah

LUCKY_DIMON [66]3 years ago
4 0
C) oxygen and D) glucose
Photosynthesis use carbon dioxide and energy from the Sun to make sugar molecules (glucose) and oxygen. Respiration use oxygen and glucose to synthesize energy-rich carrier molecules, such as ATP, and carbon dioxide is produced as a waste product.
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Answer:

E_{cell}= +ve, reaction is spontaneous

E_{cell}= -ve, reaction is non spontaneous

E_{cell}= 0, reaction is in equilibrium

Case 1: when copper metal is combined with aqueous zinc sulfate.

Cu+ZnSO_4\rightarrow Zn+CuSO_4

Here copper is undergoing oxidation ad thus acts as anode and zinc is undergoing reduction , thus acts as cathode.

E^o_{cell} = standard electrode potential =E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0_{[Cu^{2+}/Cu]}= +0.34V

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E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Cu^{2+}/Cu]}

E^0=-0.76-(+0.34)=-1.10V

Thus as E_{cell} is negative , the reaction is non spontaneous.

Case 2: when zinc metal and aqueous copper sulfate solution are combined.

Zn+CuSO_4\rightarrow Cu+ZnSO_4

Here zinc is undergoing oxidation ad thus acts as anode and copper is undergoing reduction , thus acts as cathode.

E^o_{cell} = standard electrode potential =E^0_{cathode}- E^0_{anode}

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E^0_{[Cu^{2+}/Cu]}= +0.34V

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E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}

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