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svp [43]
3 years ago
15

In simple summary form, the input of the Calvin cycle for every product is __________, and the output is __________ used for glu

cose synthesis. View Available Hint(s) In simple summary form, the input of the Calvin cycle for every product is __________, and the output is __________ used for glucose synthesis.
3 CO2 ...;
a three-carbon molecule 6 CO2 ...;
glucose ADP ... ATP NADP+ ...;
NADPH
Chemistry
1 answer:
Mumz [18]3 years ago
4 0

Answer:

3 CO₂ ......a three-carbon molecule

Explanation:

  • Photosynthesis involves the synthesis of food by green plants and algae in the presence of energy from the sun, water, and carbon dioxide.
  • The process occurs in two stages known as the light-dependent reactions and light-independent reactions also known as the Calvin cycle.
  • During light reactions of photosynthesis, water molecules are split by energy from sunlight to give out oxygen gas and hydrogen ions which are channeled to the Calvin cycle.
  • Additionally, ATP and NADPH molecules are released.
  • In the Calvin cycle, 3 molecules of carbon dioxide are combined with hydrogen ions to yield three carbon-molecules.
  • The process also uses ATP and NADPH from the light reactions.
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The rate constant for a certain reaction is k = 4.70×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the
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Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is  a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

-\frac{d[A]}{dt} =kt

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

-\frac{d[A]}{dt} =kt

\Rightarrow -\frac{d[A]}{[A]}=kdt

Integrating both sides

\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀]  = k.0 + c

⇒c= -ln[A₀]  

Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]=  0.375

⇒[A] = 0.686

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

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