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4 years ago

How much ATP is produced from a single glucose molecule in each chemical pathway?

Chemistry

2 answers:

vagabundo [1.1K]4 years ago

6 0

Glycolysis yields 2 ATP molecules, Kreb's cycle yields 2 ATP molecules, ETS yields 34 ATP, molecules

AURORKA [14]4 years ago

3 0

<h2>Answer : 38 ATP molecules</h2><h3>Explanation :</h3>

After each cellular respiration cycle there are ATP molecules obtained after oxidizing each glucose molecule

Out of which 2 ATP molecules are obtained from glycolysis cycle, 2 ATP molecules from the Krebs cycle, and about 34 ATP molecules from the electron transport system.

So, in all 2 + 2 + 34 = 38 ATP molecules.

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Need help ASAP ( will give 50 points)

Alina [70]

Answer:

Your answer would be D, Hope this helps.

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3 years ago

If an industrial worker knows that his chemical process is 87.3% efficient and he needs to

Dmitrij [34]

I legitimately think it's 87.3 grams

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3 years ago

Sunflower oil contains 0.080 mol palmitic acid (C16H32O2)/mol, 0.060 mol stearic acid (C18H36O2)/mol, 0.27 mol oleic acid (C18H3

Cerrena [4.2K]

Answer:

$x_H=0.882$

$x_N=0.118$

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: $C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)$

Linoleic: $C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)$

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

<u>Base of caculation: 1 mol of sunflower oil</u>

For oleic acid: $n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}$

$n_{Holeic}=frac{90.45 mol H_2}{hr}$

For linoleic acid: $n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}$

$n_{Holeic}=frac{395.3 mol H_2}{hr}$

$n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}$

$n_{Htotal}=frac{485.75 mol H_2}{hr}$

Applying the excess:

$n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol$

Nitrogen: $n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}$

$n_N= 42.2 mol N$

<u>After the reactions</u>:

$n_H=801.48 mol-485.75mol=315.73 mol$

and the nitrogen is inert.

Purge stream:

$n_total=42.2+315.73 mol=357.93 mol$

$x_H=\frac{315.73mol}{357.93mol}=0.882$

$x_N=\frac{42.2mol}{357.93mol}=0.118$

4 0

3 years ago

ANSWER QUICK

Novosadov [1.4K]

Answer:

1.8

Explanation:

8 0

3 years ago

Read 2 more answers

Calculate the standard entropy of vaporization of ethanol at its boiling point 285 K. The standard molar enthalpy of vaporizatio

Ivanshal [37]

Answer:

standard entropy of vaporization of ethanol = 142.105 J/K-mol

Explanation:

given data

enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × $10^{3}$ J/mol

entropy of vaporization of ethanol boiling point = 285 K

to find out

standard entropy of vaporization of ethanol

solution

we get here standard entropy of vaporization of ethanol that is expess as

standard entropy of vaporization of ethanol ΔS = $\frac{\Delta H}{T}$ .............1

here ΔH is enthalpy of vaporization of ethanol and T is temperature

put value in equation 1

standard entropy of vaporization of ethanol ΔS = $\frac{40.5*10^3}{285}$

standard entropy of vaporization of ethanol = 142.105 J/K-mol

3 0

4 years ago

Read 2 more answers