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Len [333]
3 years ago
14

Where is the energy in the products of photosynthesis?

Chemistry
2 answers:
Musya8 [376]3 years ago
8 0

The correct answer is <u><em>D. The bonds of a glucose molecule store chemical energy.</em></u> Animals need energy from plants, which have glucose in them that stores energy so when a herbivore eats the plant they get the energy from the plant.

Hope it helps and have a wonderful day/night!

serious [3.7K]3 years ago
3 0
The bonds of a glucose molecule store chemical energy
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how does photosynthesis and respiration in plants influence the amount of co2 in the atmosphere during a year?
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An increase in the amount of CO2 well increase the rate of photosynthesis, Carbon dioxide concentration will directly affect the rate of photosynthesis as it is used in the photosynthesis reaction.Increased amount of CO2 will increase the rate of photosynthesis to a certain limit, after which a further increase in its amount will no longer increase the rate any further.

Hope this helps.

Cheers! :)

8 0
3 years ago
Y'all what about this one
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A=8, B=7,C=N

D=26, E=30, F =Fe

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3 years ago
Analyze the map. How would the temperature on the east coast of South America (near Brazil) differ from the west coast of South
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5 0
2 years ago
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For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl
Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of H_2S is 3.92  atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

      2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}

On reversing the reaction:

     2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

initial pressure  4.00atm    2.00 atm       0

eqm          (4.00-2x)atm      (2.00-x) atm      2x atm

K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}

K_p'=\frac{1}{K_p}=0.67\times 10^5

2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)

0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}

x=1.96

[H_2S]=2x=2\times 1.96=3.92 atm

Thus approximate equilibrium partial pressure of H_2S is 3.92 atm

3 0
3 years ago
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