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Nitella [24]
3 years ago
13

PLEASE HELP WITH CHEMISTRY !! BRAINLIEST AND 10 POINTS !!

Chemistry
1 answer:
NeTakaya3 years ago
7 0

Answer:

c is your answer there u go

Explanation:

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The free energy for the oxidation of glucose to CO2 and water is -686 kcal/mol, and the free energy for the reduction of NAD+ to
vitfil [10]

Answer:

Most of free energy available from oxidation of the glucose remains in pyruvate.

Explanation:

The overall reaction of the process glycolysis is:

Glucose + 2 NAD⁺ + 2 ADP + 2 Pi ⇒ 2 Pyruvate + 2 NADH + 2 H⁺ + 2ATP

Glucose is oxidized to give 2 molecules of pyruvate and 2 molecules of NADH and ATP (Energy currency).

<u>Though the free energy of oxidation of glucose is high but only 2 NADH is formed because the most of the free energy that is being released from the oxidation of glucose remains in the pyruvate which is produced in the reaction and thus only 2 molecules are formed.</u>

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Describe how you created the 0.075 M glucose solution from the 1 M stock solution. Show your calculations and give details about
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by addind

Explanation:idk how to do this

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Identify one step in the water cycle that is driven by the force of gravity
iren2701 [21]

Answer:

i believe its precipitation??

Explanation:

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What type of bond is present between two atoms of tungsten in a light bulb filament?
stiks02 [169]

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Explanation:

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8 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
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