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Crazy boy [7]
3 years ago
6

In most cells, not all of the carbon compounds that participate in glycolysis and the citric acid cycle are converted to carbon

dioxide by cellular respiration. What happens to the carbon in these compounds that does not end up as CO2? See Concept 9.6 (Page 183)
Chemistry
1 answer:
Dmitrij [34]3 years ago
4 0

The carbon compounds are removed from the processes to serve as building blocks for other complex molecules.

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This is agree or disagree I need help ASAP!!
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Answer:

disagree

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Explanation:

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How many molecules of F2 in 90g F2
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3 years ago
I WILL MARK YOU BRAINLIST PLEASE HELP
tekilochka [14]

Answer:  See below

Explanation:  a.  The mass of an element is composed of:

protons:     1 amu each

neutrons:   1 amu each

electrons:   0 amu each

Only the protons and neutrons are counted in the atomic mass of an element

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7 0
2 years ago
In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.
Mamont248 [21]

Answer:

Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let \rm HA denote this acid.

\rm HA \rightleftharpoons H^{+} + A^{-}.

Initial concentration of \rm HA without any dissociation:

[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}.

After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}.

Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

5 0
3 years ago
PLEASE HELP
dalvyx [7]

It has a -1 charge, so it would gain one!

5 0
3 years ago
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