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creativ13 [48]
2 years ago
8

2. Which of the following diagrams accurately represents the use of gases in both cellular respiration

Chemistry
2 answers:
sesenic [268]2 years ago
7 0

Answer:

In algae or any other photosynthetic organism, there is the internalization of carbon dioxide gas in the reaction with water in the presence of sunlight to produce sugar molecules and oxygen gas in the atmosphere. So, photosynthesis is the process that releases or produces oxygen gas.

In cellular respiration (aerobic) the product of photosynthesis, glucose molecules, and oxygen react to produce the energy and releases carbon dioxide and water as byproducts.

astra-53 [7]2 years ago
4 0

Answer:

A, Cellular respiration gives Co2 to photosynthesis and Photosynthesis gives O2 to cellular respiration.

Explanation:

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Answer : The correct option is, (A) -101.37 KJ

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First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

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Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 87ml+87ml=174ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

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q = heat absorbed = ?

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T_{final} = final temperature of water = 317.4 K

T_{initial} = initial temperature of metal = 298 K

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

q=174g\times 4.18J/g^oC\times (317.4-298)K

q=14110.008J=14.11KJ

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole

Therefore, the enthalpy of neutralization is, -101.37 KJ

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