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valina [46]
3 years ago
13

Which of the following are products of the light-dependent reactions of photosynthesis that are utilized in the light-independen

t reactions?
ATP and NADPH
water and oxygen
oxygen and glucose
carbon dioxide and glucose
Chemistry
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

\boxed {\tt Oxygen \ and \ glucose}

Explanation:

The photosynthesis equation is:

energy+6CO₂+6H₂O--> C₆H₁₂O₆+6O₂

The reactants are light energy, carbon dioxide, and water. The products are <u>glucose and oxygen.</u>

<u />

The products of photosynthesis become the products of the light-independent reaction called cellular respiration, which creates useable energy. Therefore, the correct answer is oxygen and glucose.

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Answer: PV = nRT

A gas at STP... This means that the temperature is 0°C and pressure is 1 atm.

R is the gas constant which is 0.08206 L*atm/(K*mol)

Rearranging for volume

V = nRT/P

The temperature and number of moles are held constant. This means that this uses Boyle's Law. (The ideal gas law could be manipulated to give us this result when T and n are held constant.)

PV = k

where k is a constant.

This means that

P₁V₁ = k = P₂V₂

P₁V₁ = P₂V₂

(1 atm) * (1 L) = (2 atm) * V₂

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A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
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Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

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3 years ago
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