Nitrous acid, hno2, has an acid dissociation constant - ka of 7. 1 ✕ 10-4. what are [h3o ], [no2-], and [oh -] in 0. 40 m hno2 - 4829 M [OH^-] = 1.439 x 10^-14 M
The acid dissociation constant (Ka) is used to differentiate between strong and weak acids. Strong acids have very high Ka values. The Ka value is determined by examining the equilibrium constant for acid dissociation. The acid dissociates more readily as the Ka increases.
The original molecular definition of an acid, according to Arrhenius, is a molecule that dissociates in an aqueous solution, releasing the hydrogen ion H+ (a proton): HA A + H+. acid dissociation constant is an equilibrium constant for this dissociation reaction.
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Answer:
(a) 
(b) 
(c) 
Explanation:
Hello,
(a) In this case, since entropy remains unchanged, the constant
should be computed for air as an ideal gas by:


Next, we compute the final temperature:

Thus, the work is computed by:

(b) In this case, since
is given, we compute the final temperature as well:

And the isentropic work:

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

Regards.
Answer:
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Answer:
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they are all metals and that most of them are hard, strong, and lustrous, have high melting and boiling points, and are good conductors of heat and electricity.