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Nana76 [90]
2 years ago
7

Which of the following shows the conservation of mass during cellular respiration? 3 CO2 3 H2O energy → 3 C6H12O6 3 O2 6 CO2 6 H

2O energy → C6H12O6 6 O2 C6H12O6 6 O2 → 6 CO2 6 H2O energy 6 H2O C6H12O6 → 6 O2 energy.
Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
4 0

The reaction that has been, following the law of conservation of mass has been \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy.

The law of conservation has been given in the chemical reaction that there has been no loss or gain of the mass and energy.

The law of conservation has been evident when there has been an equal number of atoms of each element on the product and the reactant side.

<h3 /><h3>Conservation of mass in Cellular respiration</h3>

The following reactions have been identified as:

  • \rm 3\;CO_2\;+\;3\;H_2O\;+\;Energy\;\rightarrow\;3\;C_6H_1_2O_6

Carbon atoms

Reactant = 3

Product = 18

Oxygen atoms

Reactant = 9

Product = 18

Hydrogen atoms

Reactant = 6

Product = 36

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

  • \rm 3\;O_2\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Energy\;\rightarrow\;C_6H_1_2O_6\;+\;6\;O_2

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 24

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

  • \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy

Carbon atoms

Reactant = 6

Product = 6

Oxygen atoms

Reactant = 18

Product = 18

Hydrogen atoms

Reactant = 12

Product = 12

The number of atoms is equal on the product and reactant side, thus follows the law of conservation of mass.

  • \rm 6\;H_2O\;+\;C_6H_1_2O_6\;\rightarrow\;6\;O_2\;+\;Energy

Carbon atoms

Reactant = 6

Product = 0

Oxygen atoms

Reactant = 12

Product = 12

Hydrogen atoms

Reactant = 24

Product = 0

The number of atoms is not equal on the product and reactant side, thus not follows the law of conservation of mass.

Thus, the reaction that has been following the law of conservation of mass has been \rm C_6H_1_2O_6\;+\;6\;O_2\;\rightarrow\;6\;CO_2\;+\;6\;H_2O\;+\;Energy.

Learn more about law of conservation, here:

brainly.com/question/2175724

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What is the ph of a 0.10 m solution of ethylamine at 25 oc? the pkb of ethylamine at 25 ºc is 3.25. ethylamine is a weak base?
denis-greek [22]
To determine the pH of a weak base, we use the equation:

pH = 14 + 0.5 log Kb

Therefore, 


pH = 14 + 0.5 log 3.25
pH = 14.26

A weak base is a base which does not fully dissociates into ions when in solution. The solution would contain cations, anions and the compound itself. Hope this helps.
5 0
3 years ago
What is the density of water if you have 50.0 grams of water and a volume of 50.0 millimeters
Gnom [1K]

Answer:

\boxed {\tt 1.0 \ g/mL}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass of the water is 50.0 grams.

The volume of the water is 50.0 milliliters.

m= 50.0\ g \\v=50.0 \ mL

Substitute the values into the formula.

d=\frac{50.0 \ g}{50.0 \ mL}

Divide.

d= 1.0  \ g/mL

The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³

8 0
3 years ago
CAN ANYONE ANSWER THIS ONE CHEMISTRY QUESTION PLZZZ!!!
xz_007 [3.2K]

Answer:

2.8 oxygen

Explanation:

8 0
3 years ago
Read 2 more answers
"calculate the ratio of the velocity of hydrogen molecules to the velocity of carbon dioxide molecules at the same temperature"
Ne4ueva [31]

Answer: 1:4.69

Explanation:

The ratio can be expressed as:

Ua/Ub= √(Mb/Ma)

Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2

Mb is the molecular mass of CO2 = 44

Therefore

Ua/Ub= √(44/2)

Ua/Ub = 4.69

Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69

which implies hydogen is about 4.69 times faster than carbon dioxide.

8 0
4 years ago
Q2.  0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht
Vera_Pavlovna [14]

Answer:

<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>

Explanation:

No. of moles = \frac{mass}{molar mass}

= \frac{0.254}{204} = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= \frac{0.001245}{0.02} = 0.062 mol/L

We know that:

C_{1} V_{1} = C_{2} V_{2}

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>

<u><em>= 2.49g/L.</em></u>

3 0
3 years ago
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