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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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erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:

  • 23.8 °C → 23.8 + 273.15 = 296.95 K
  • 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * \frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}} = 7.85 * 10⁻³ mol SO₂ = n

  • Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

  • Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0
3 years ago
What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and N
Alecsey [184]

Answer:

Cr(OH)2(s), Na+(aq), and NO3−(aq)

Explanation:

Let is consider the molecular equation;

2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)

This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.

Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.

4 0
4 years ago
You have a 250. -ml sample of 1. 28 m acetic acid (ka = 1. 8 x´ 10–5). calculate the ph of the best buffer.
Nadya [2.5K]

The ph of the best buffer is 4.74

The given acetic acid is a weak acid

The equation of the pH of the buffer

pH = pKa + log ( conjugate base / weak acid ).

For best buffer the concentration of the weak acid and its conjugate base is equal.

pH = pKa + log 1

pH = pKa + 0

pH = pKa

given Ka = 1.8 × 10⁻⁵

pKa = - log ka

pH = -log ( 1.8 × 10⁻⁵ )

pH = 4. 74

Hence the pH of the best buffer is 4.74

Learn more about the pH on

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4 0
2 years ago
Read 2 more answers
Which of the following equations can be used to calculate pOH?
Ivenika [448]
POH = 14 - pH
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6 0
3 years ago
Assuming that a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, determine how many kg of co2 are prod
sleet_krkn [62]

Answer: -

If a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, 0.26 kg of CO₂ are produced for each tank of gasoline burned.

Explanation: -

Density of the gasoline = 0.77 kg / liter

Volume of the tank containing the gasoline = 80 liter.

Mass of gasoline produced from each tank

= Volume of the tank containing the gasoline x Density of the gasoline

= \frac{0.77 kg}{1 liter} x 80 liter

= 61.6 kg

Chemical formula of gasoline = C₈H₁₈

Molar mass of gasoline C₈H₁₈ = 12 x 8 + 1 x 18 = 114 g/ mol

Number of moles of C₈H₁₈ = \frac{61.6 g}{114 g} x 1 mol

= 0.54 mol of C₈H₁₈

The chemical equation for the burning of gasoline is

2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

From the balanced equation we see

2 mol of C₈H₁₈ gives 16 mol of CO₂

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= 4.32 mol of CO₂

Molar mass of CO₂ = 12 x 1 + 16 x 3 =60 g / mol

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=\frac{60g x 4.32 mol}{1 mol}

= 259.2 g

= \frac{259.2}{1000}

= 0.259 Kg

= 0.26 kg rounded off to 2 significant figures.

Thus if a tank of gasoline contains 80 liters and that its density is 0.77 kg/liter, 0.26 kg of CO₂ are produced for each tank of gasoline burned.

4 0
3 years ago
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