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rosijanka [135]
3 years ago
6

Calculate the amount of ATP in kg that is turned over by a resting human every 24 hours. Assume that a typical human contains ~5

0g of ATP (Mr 505) and consumes ~8000 kJ of energy in food each day. The energy stored in the terminal anhydride bond of ATP under standard conditions is 30.6 kJmol-1. Assume also that the dietary energy is channeled through ATP with an energy transfer efficiency of ~50%.
Chemistry
1 answer:
valentinak56 [21]3 years ago
5 0

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

moles = mass/molecular mass

The molecular mass of ATP is 507.18 g per mol

Now by putting the values we get,  

mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

= Total production of ATP + Total ATP available

= 66.298 kg + 0.05 kg

= 66.348 kg

Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.  

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

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Explanation:

Gas mixture:

n_{Ar}= 4 mol

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n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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3 years ago
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Answer:

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