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zysi [14]
3 years ago
15

A cell has lost the ability to release energy from glucose for the cell to use. In which of the following cell structures would

this energy conversion be performed?
A nucleusnucleus

B mitochondriamitochondria

C cell membranecell membrane

D cytoplasm
Chemistry
2 answers:
Irina-Kira [14]3 years ago
7 0

Answer:

b

Explanation:

b? i think

soldier1979 [14.2K]3 years ago
4 0
B) mitochondria
This is because the mitochondria is the power house of the cell. And energy conversion happens in the mitochondria
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What kind of change is a change in state of matter and why<br> Physical or chemical
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Answer:

physical

Explanation:

Common changes of state include melting, freezing, sublimation, deposition, condensation, and vaporization. I hope this helps. :)

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What is a food web made up of
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Answer:

A food web is made up of all the food chains in a single ecosystem. Each living thing in an ecosystem belongs to many food chains. A food chain is a path that energy takes through a certain ecosystem. Trophic Levels. Organisms in food webs are grouped into categories called trophic levels.

Explanation:

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2 years ago
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A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

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