To determine the correct sequence of acidic strength for the given aliphatic carboxylic acids, let's analyze each compound:
The electron-donating effect of alkyl groups decreases the acidity because they destabilize the carboxylate ion formed after deprotonation. Thus, the more extended the alkyl chain, the weaker the acid.
Based on the analysis above, the sequence of acidic strength in decreasing order is:
Therefore, the correct answer is: \(HCOOH > CH_3COOH > CH_3CH_2COOH > CH_3CH_2CH_2COOH\).
The acidic strength of carboxylic acids is influenced by the electron-withdrawing or electron-donating effects of alkyl groups:
HCOOH (formic acid) is the most acidic as it has no electron-donating alkyl group, which would reduce acidity.
CH$_3$COOH (acetic acid) is less acidic because the methyl group (CH$_3$) is weakly electron-donating.
CH$_3$CH$_2$COOH (propionic acid) is even less acidic due to the larger electron-donating ethyl group.
CH$_3$CH$_2$CH$_2$COOH (butyric acid) is the least acidic because the longer alkyl chain has a stronger electron-donating effect.
The correct order of acidic strength is:
\[ \textbf{HCOOH > CH$_3$COOH > CH$_3$CH$_2$COOH > CH$_3$CH$_2$CH$_2$COOH}. \]
Consider the following reaction of benzene. the percentage of oxygen is _______ %. (Nearest integer) 





MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
