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3 years ago

The pentose phosphate pathway (PPP) is localized to the cytosol.

1 answer:

5 0

Answer: The pentose phosphate pathway (PPP) is localized to the cytosol because fatty acid synthesis uses the NADPH generated by the PPP.

Explanation:

The pentose phosphate pathway is mainly catabolic and provides an alternative glucose oxidizing pathway for the generation of NADPH that is required for reductive biosynthetic reactions such as those of cholesterol biosynthesis, bile acid synthesis, steroid hormone biosynthesis, and fatty acid synthesis.

Fatty acid biosynthesis occurs in the cytosol and requires the reducing equivalent NADPH in large amounts. The main source of generating NADPH in animal cells, the pentose phosphate pathway is therefore, localized in the cytosol in order to furnish a strongly reducing environment for fatty acid biosynthesis to proceed.

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2Fe(s) +3H2SO4(aq) →Fe2(SO4)3(aq) +3H2(g)When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, what is the percent y

Elden [556K]

Answer:

1040%

Explanation:

To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:

Percent yield = Actual yield (5.40g) / Theoretical yield * 100

Moles Fe -Molar mass: 55.845g/mol-:

10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.

For a complete reaction of these moles there are necessaries:

0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.

As there are 14.8 moles of the acid, Fe is limiting reasctant.

The moles of H2 produced are:

0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2

The mass is:

0.277 moles H2 * (2.016g/mol) = 0.558g H2

Percent yield is:

5.40g / 0.558g * 100 = 1040%

It is possible the experiment wasn't performed correctly

7 0

2 years ago

Please select the word from the list that best fits the definition

AveGali [126]

Answer:

I choose D option because may be it's correct

3 0

2 years ago

Read 2 more answers

At 580 k, 6. 42% of the molecules are in the chair form. calculate the value of the equilibrium constant for the reaction as wri

Ann [662]

Equilibrium chemical reaction between chair and boat forms is:
Kc = $\frac{[Boat form]}{[Chair form]}$
Suppose the total number of molecules is x
Number of molecules of chair form is 6.42/100 x
Number of molecules of boat form is 93.58/100 x
Kc = $\frac{(0.9358x)}{(0.0642x)}$ = 14.6

5 0

2 years ago

How many hydrogen atoms are in 5 molecules of isopropyl alcohol

pychu [463]

You have to double them

5 0

3 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0

3 years ago