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Olin [163]
3 years ago
10

Specific amino acids that line the selectivity filter of the K channel can dehydrate K and allow passage of that ion but cannot

coordinate the dehydration of Na . True or False
Chemistry
1 answer:
Bogdan [553]3 years ago
5 0

Answer:

Hi

True

Explanation:

Since on the inner and outer surfaces of the membrane, a series of negatively charged amino acids are found, which increase the local concentration of cations. The path of the ions begins on the inner surface filled with water molecules where the ion can retain its hydration sphere. Two thirds of its interior in the membrane the inside of the channel narrows in the region of the selectivity filter, forcing the ion to separate from the water molecules. Oxygen atoms in the selectivity filter replace the water molecules in the K+ hydration sphere, forming a series of coordination spheres through which the ion moves. The preferential stabilization of K+ against Na+ is the basis of the ion selectivity of this filter.

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\large \boxed{\text{-41.2 kJ/mol}}

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\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

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\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

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