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saul85 [17]
3 years ago
5

All of the cells in the body need oxygen. Hemoglobin molecules in red blood cells transport oxygen through the bloodstream. Oxyg

en is loaded onto hemoglobin molecules in the lungs and unloaded from the hemoglobin molecules in the tissues.
What drives the unloading of oxygen from hemoglobin molecules in the tissues?

A. the low partial pressure of oxygen in the tissues
B. the high partial pressure of carbon dioxide in the tissues
C. the low partial pressure of carbon dioxide in the tissues
D. the high partial pressure of oxygen in the tissues
Chemistry
1 answer:
Yakvenalex [24]3 years ago
5 0

Answer: Option B) the high partial pressure of carbon dioxide in the tissues.

Explanation:

The high concentration of oxygen in the alveoli of the lungs unloads H+ and carbon dioxide (CO2) from the hemoglobin, just as the high concentration of H+ and CO2 in active tissues unloads Oxygen from hemoglobin molecules.

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A chemist measures the enthalpy change ?H during the following reaction: Fe (s) + 2HCl (g) ? FeCl2 (s) + H2 (g) =?H?157.kJ Use t
erik [133]

Correct Question:

A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol

Answer:

-314 kJ

+628 kJ

+157 kJ

Explanation:

The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.

If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.

1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)

This reaction is the product of the given reaction by 2, so

ΔH = 2*(-157) = -314 kJ

2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)

This reaction is the inverted reaction given multiplied by 4, so

ΔH = 4*(157) = +628 kJ

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