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3 years ago

The electrons stripped from glucose in cellular respiration end up in which compound?

Chemistry

1 answer:

Inga [223]3 years ago

6 0

Answer:

The electrons stripped from glucose in cellular respiration end up in compound water by the reduction of oxygen.

Explanation:

During electron transport chain electrons are donated by various reducing equivalents such as NADH,FADH2.The donated electrons then moves through various electron carriers .

During electron transport chain oxygen(O2) act as terminal electron acceptor which accept the electron from complex 4 and thereby get reduced to form water.(H2O).

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A reaction that occurs in the internal combustion engine is N₂(g) + O₂(g) ⇄ 2NO(g) (c) What is the significance of the different

vodomira [7]

The response would become spontaneous if the value of ΔG° was negative.

According to the estimated value of ΔG°, it is shown that ΔG° value decreases as temperature value increases. The value shifts from being more favorable to being less favorable. It would appear that the value of ΔG° would be negative at a specific temperature, causing the reaction to occur spontaneously.

The reaction is in an equilibrium state if ΔG = 0. If ΔG < 0, the reaction is spontaneous in the direction written. The relationship between terms from the equilibrium is paralleled by the relevance of the sign of a change in the Gibbs free energy.

Learn more about ΔG° here:

brainly.com/question/14512088

#SPJ4

8 0

1 year ago

A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C

Mamont248 [21]

Answer:

$T_{eq}=23.85^oC$

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

$\Delta H_{Cu}=-\Delta H_{H_2O}$

Therefore the equilibrium temperature shows up as:

$m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\$

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

$T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC$

Best regards.

6 0

3 years ago

Read 2 more answers

Which of the following is a synthesis reaction?

zhuklara [117]

The last one is the answer to your question. D is the answer

6 0

3 years ago

Read 2 more answers

Which ion in the ground state has the same electron configuration as an atom of neon in the ground state?

Ainat [17]

Cl-

Lets see how

Z for Cl is 9
One electron is added so new Z=10

Electronic configuration

1s²2s²2p⁶

[Ne]

option B is correct

6 0

2 years ago

Read 2 more answers

For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. The reaction is endothermic. What do yo

jeyben [28]

Answer:

The correct answer is : 'the concatenation of NO will increase'.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs.

$2NOBr(g)\rightleftharpoons 2NO(g) + Br_2(g)$

As, this is an endothermic reaction, increasing temperature will add more heat to the system which move equilibrium in the forward reaction with decrease in temperature. Hence, the equilibrium will shift in the right direction.

So, the concatenation of NO will increase.

7 0

4 years ago