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xz_007 [3.2K]
3 years ago
7

Bottom two answer pls

Chemistry
2 answers:
Alex17521 [72]3 years ago
3 0

Answer:

i cant see it

Explanation: for real

Marysya12 [62]3 years ago
3 0

Glucose is ‘raw material’ in cellular respiration. It passes through the glycolysis cycle, in the cytoplasm of the cell. This occurs in virtually all cells. In eukaryotic cells, with mitochondria, the product of glycolysis (usually pyruvate) is channeled to the organelle and pass through the Citric acid /Krebs cycle that produces even more ATPs than glycolysis.

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The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8
scZoUnD [109]

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane =\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M

The given balanced equilibrium reaction is,

                            cyclohexane  ⇔  methylcyclopentane

Initial conc.                 0.025 M           0

At eqm. conc.       (0.025-x)M       (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

Now put all the given values in this expression, we get :

0.12=\frac{(x)}{(0.025-x)}

By solving the term 'x', we get :

x =  0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

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