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Lemur [1.5K]
3 years ago
6

Living things respire to stay alive. In the mitochondria, sugar is combined with oxygen to produce energy, water and carbon diox

ide. Write the equation of this chemical reaction
Chemistry
2 answers:
olya-2409 [2.1K]3 years ago
8 0

Answer:

C6H12O6 + 6O2 → 6CO2 + 6H2O

Explanation:

Cellular respiration occurs in the mitochondria of cells. It is a process in which  sugar is combined with oxygen to produce energy, water and carbon dioxide. This is the major process by which energy is released in living organisms.

Aerobic respiration involves a series of chemical reactions.  These reactions commence with sugar and oxygen then it produces  carbon dioxide and water according to the reaction equation; C6H12O6 + 6O2 → 6CO2 + 6H2O.

Reika [66]3 years ago
8 0

Answer: The equation for this Chemical reaction is:

C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) + heat (energy)

Explanation:

Mitochondria is the site of chemical energy conversions for cell activities. In the mitochondria, sugar (chemical energy) is combined with oxygen to produce energy, water and carbon dioxide. The equation for this reaction is already given above.

All living cells need energy for the metabolic processes essential for life. They get this energy from the chemical energy stored in food (sugar).

In this process, oxygen is necessary for the chemical energy in sugar to be converted to carbondioxide, water and release of heat energy. It is therefore an oxidative reaction.

Furthermore, energy is not released in one big step as shown in the above reaction, but in a series of small steps which are catalysed by enzymes. The energy that is released, bit by bit, is stored in adenosine triphosphate (ATP) molecules. The oxidation of glucose to release energy in this manner is known as CELLULAR RESPIRATION. It occurs also in the mitochondria of all living cells.

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Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.
Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

ln(K)=\frac{\Delta _RG}{RT}

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

B. decrease, since less reactant is forming the products.

Best regards.

8 0
4 years ago
Read 2 more answers
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

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

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
3 years ago
If the conjugate base of a molecule has a pkb of 1.4, what would you expect the molecule to be?
maw [93]

If the conjugate base of a molecule has a pKb of 1.4, the molecule should be a Weak Acid.

Notice this question gives us the pKb of the molecule, not the pKa. Because of this, the pH scale basically gets reversed, so lower numbers in pKb correlate with stronger bases, and higher numbers in pKb correlate with stronger acids - the exact opposite of the pH scale.

It's important to make sure you completely understand the terms of conjugate base, conjugate acid, pKb, pKa, and how they all relate. It's easy to mix up the meanings of these definitions.

Here are the two other pieces of information you need to know to correctly answer this question:

  • Strong acids have a weak conjugate base.
  • Strong bases have a weak conjugate acid.

So if the problem says you have a strong conjugate base, then the molecule must be a weak acid. To illustrate this, think of ammonium, NH4+. Ammonium is a weak acid, but the conjugate base of ammonium is ammonia, NH3, which is a reasonably good base.

Learn more about conjugate base here : brainly.com/question/22514615

#SPJ4

3 0
2 years ago
How does a filter separate mixtures like sand and water?​
Talja [164]

Answer:

Filtration is a method for separating an insoluble solid from a liquid. When a mixture of sand and water is filtered: the sand stays behind in the filter paper (it becomes the residue ) the water passes through the filter paper (it becomes the filtrate )

Explanation:

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3 years ago
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Naya [18.7K]

Answer:

A B D

Explanation:

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