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Natali [406]
3 years ago
10

When a plant is entering the Calvin cycle of photosynthesis_____.

Chemistry
1 answer:
Andrei [34K]3 years ago
4 0

B). light energy is not required to proceed

Explanation:

In the Calvin cycle of photosynthesis, light energy is not required. The Calvin cycle is light independent and it is made up of a series of redox reactions.

  • During photosynthesis reactions, green plants manufacture their food using carbon dioxide, sunlight and water.
  • During the Calvin cycle aspect, light energy is not required for chemical reactions to take place. The light energy helps to move electrons.
  • The cycle is also known as dark reactions.
  • It is at this stage that carbon dioxide combines with water to form glucose.
  • The reaction is initiated with light energy which produces NADPH and ATP.
  • The Calvin cycle follows by using the NADPH and ATP to produce glucose in the dark phase.

Learn more:

ATP brainly.com/question/2953868

Light dependent reactions brainly.com/question/6866300

#learnwithBrainly

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What is the correct answer?!????
ddd [48]

Answer:

B

Explanation:

Dalton worked with mainly about the chemistry of atoms.

how do atoms combine to form various molecules.

—rather than the details of the physical, internal structure of atoms, although he never denied the possibility of atoms' having a substructure.

4 0
3 years ago
How can we make use of acids or bases to Prevent the growth of microorganisms in swimming pools?
pentagon [3]
By putting smarticle particles 
8 0
3 years ago
An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.
aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

6 0
3 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
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stiks02 [169]

spindle fibers

i think the next is two nuclei and still together but that stage is not anaphase. Anaphase is when the sister chromatids are pulled apart

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