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MrRa [10]
3 years ago
5

En el aire que respiramos y en el aire comprimido de los tanques de los buzos se encuentra, además de oxígeno, el gas nitrógeno

(N2). Este gas entra y sale de los pulmones sin consecuencias para el organismo pero, a más de 30 m de profundidad del agua, este gas se disuelve en el torrente sanguíneo.
Cuando los buzos suben a la superficie abruptamente, la presión disminuye, lo que provoca que se reduzca la solubilidad del nitrógeno disuelto y en consecuencia este gas salga de la sangre en forma violenta. Este proceso puede provocar problemas de salud o incluso la muerte del buzo, por lo cual se recomienda que suban lentamente y haciendo pausas.
a) ¿Es la solubilidad una propiedad intensiva o extensiva?
Chemistry
1 answer:
marysya [2.9K]3 years ago
7 0

Answer: can someone please translate so i can answer.

Explanation:

sorry

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Sort the following chemicals for their ability to act as an oxidizer. List from strongest to weakest; 1 being strongest, 6 being
spin [16.1K]
Answer : If we list the given chemicals according to their increasing oxidising ability then the order will be like this; 1 being the strongest and 6 being the weakest 
1. K > 2. Ca >3. Ni> 4. Cu> 5. Ag> 6.Au

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7 0
3 years ago
Read 2 more answers
Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th
WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0
3 years ago
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