Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
equilibrium
Heat transfers from a body with high temperature to a body with low temperature until both bodies are in the same temperature.
Explanation:
As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
..........(1)
..............(2)
The final reaction is as follows:
.............(3)
Therefore, adding (1) and (2) we get the final equation (3) and value of
at 298 K will be as follows.
=
+
= -314 kJ + (-80) kJ
= -394 kJ
Thus, we can conclude that
at 298 K for the given process is -394 kJ.
Answer:
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Since monosaccharides are the simplest units of carbohydrates, then they are made up of carbon, hydrogen, and oxygen atoms.
empirical formula : CxH2xOx
where

Monosaccharides can be classified by the number x of carbon atoms they contain: triose(3) tetrose (4), pentose (5), hexose (6), heptose (7), and so on.