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Sladkaya [172]
3 years ago
7

Assume that the complete combustion of one mole of glucose to carbon dioxide and water liberates 2870 kJ/mol2870 kJ/mol ( Δ????°

′=−2870 kJ/molΔG°′=−2870 kJ/mol ). If one contraction cycle in muscle requires 67 kJ67 kJ , and the energy from the combustion of glucose is converted with an efficiency of 39%39% to contraction, how many contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose? Round your answer to the nearest whole number.
Chemistry
1 answer:
MaRussiya [10]3 years ago
5 0

Answer:

Number of contraction cycles that could theoretically be fueled by the complete combustion of one mole of glucose is around 17

Explanation:

Energy released during the complete combustion of 1 mole glucose = 2870 kJ

Energy required/muscle contraction cycle = 67 kJ/contraction

Energy conversion efficiency = 39%

Actual amount of energy converted to contraction per mole of glucose is:

=\frac{39}{100} *2870 kJ=1119.3 kJ

Total contraction cycles fueled by the above energy is:

=\frac{1119.3\ kJ}{67\ kJ/contraction} =16.7\ i.e.\ around\ 17\ contractions

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PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.
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i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).

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ii. Given the dissolution of some substance

xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)},

the Ksp, or the solubility product constant, of the preceding equation takes the general form

K_{sp} = [B]^y [C]^z.

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}.

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iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.

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1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?
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1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?

N2 + O2 → 2NO

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For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O

For 25 moles of HCl we need 25 moles of NaOH to produce 25 moles of NaCl and 25 moles of H2O

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Explanation:

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