MCQ
Which of the following is water soluble
- AInsulin
- Balbumins
- CPyridoxine
- ✓All of above
✓
Answer
Correct option: D.
All of above
d
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$\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
R-C-O-R'+R''OH\overset{{{H}^{\oplus }}}{\longleftrightarrow} \\
\end{matrix}\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
R-C-O-R''-R'OH \\
\end{matrix}$
→O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
R-C-O-R'+R''OH\overset{{{H}^{\oplus }}}{\longleftrightarrow} \\
\end{matrix}\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
R-C-O-R''-R'OH \\
\end{matrix}$
Above reaction is/an example of
The incorrect statement among the following is
Given below are two statements, one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
→Assertion $A$ : $\left[ CoCl \left( NH _3\right)_5\right]^{2+}$ absorbs at lower wavelength of light with respect to $\left[ Co \left( NH _3\right)_5\left( H _2 O \right)\right]^{3+}$
Reason $R$ : It is because the wavelength of the light absorbed depends on the oxidation state of the metal ion.
In the light of the above statements, choose the correct answer from the options given below :
Which of the following is used for lowering blood pressure?
Reduction of acetyl chloride with ${H_2}$ in presence of $Pd$ gives
→$\wedge^{{\circ}}_\text{m}\text{H}_2\text{O}$ is equal to $.......$
$a. \wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
$b. \wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}-\wedge^{{\circ}}_{\text{m}(\text{NaOH})}$
$c. \wedge^{{\circ}}_{\text{(HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}$
$d. \wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}$
→$a. \wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
$b. \wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}-\wedge^{{\circ}}_{\text{m}(\text{NaOH})}$
$c. \wedge^{{\circ}}_{\text{(HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}$
$d. \wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}$
In the moleucle $I_2Cl_6$, choose the $CORRECT$ statement $(s)$
→Most stable complex ion is
$Cu ( s )+ Sn ^{2+}(0.001 M ) \rightarrow Cu ^{2+}(0.01 M )+ Sn ( s )$
→The Gibbs free energy change for the above reaction at $298\, K$ is $x \times 10^{-1} \,k\,J\, mol ^{-1}$;
The value of $x$ is ..... [nearest integer]$\left [\text { Given : } E _{ Cu ^{2} / / Cu }=0.34\, V ; E _{ Sn ^{2} / Sn }^{\ominus}=-0.14 \,V ; F=96500\, C\, mol ^{-1}\right]$
Formaldehyde is used as formalin in industries: