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Marta_Voda [28]
3 years ago
8

how does the processes of photosynthesis and cellular respiration allow for the flow of energy in organisms

Chemistry
1 answer:
Naddika [18.5K]3 years ago
8 0
Photosynthesis and cellular respiration go hand and hand. Remember the two equations are just a reverse of the opposite equation. 

Cellular Respiration-
C₆H₁₂O₆ + 6O₂ (Yeilds or Makes) 6CO₂ + 6H₂O + ATP (Or Energy)

Photosynthesis-
6CO₂ + 6H₂O + ATP (Or Sunlight) (Yeilds or Makes) C₆H₁₂O₆ + 6O<span>₂  
</span>
When a plant goes through photosynthesis it produces oxygen as a waste product, which you should know is what animals use to breathe, well when animals use oxygen in the process, they also make a waste product which happens to be Carbon Dioxide, which a plant uses to make glucose during photosynthesis, so if we didin't have one we wouldn't have the other.
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sugar taken in a boiling tube is strongly heated. a) write the procedure of this experiment . b) which are the products obtained
Tpy6a [65]

Answer:

chemical changes

Explanation:

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6 0
2 years ago
If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.
Klio2033 [76]

Answer : The final temperature of the mixture is 22.7^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.3J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol

m_2 = mass of water

\rho_1 = density of ethanol = 0.789 g/mL

\rho_2 = density of water = 1.0 g/mL

V_1 = volume of ethanol = 45.0 mL

V_2 = volume of water = 45.0 mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 9.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get

(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.7^oC

Therefore, the final temperature of the mixture is 22.7^oC

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777dan777 [17]

Answer:

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Explanation:

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