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A hydrocarbon ($C_5{H}_8$) given in the following scheme performs Tollens' test.

Identify product 'C'.

A stable species $C$ can be obtained through unstable nuclides $A$ and $B$. The half life period for conversion $A\to C$ is $T$ and that for conversion $B\to C$ is $2T$. Initially the sample contains $N$ nuclides of $A,N$ nuclides of $B$ and no nuclides for $C$. After $t=4T$, the nuclides of species $C$ will be equal to

1L flask contains nitrogen along with a drop or two of water at ${40}^{\circ }C$, the total pressure being 760 torr. If all the conents are transferred to another flask of 0.5 L at the same temperature, the pressure set up in the second flask will be (aqueous tension at ${40}^{\circ }C=55torr)$

Consider the following processes and respective energy change-

$M\left(s\right)\to M\left(g\right).....\left(a\right)$

$M\left(s\right)\to M^{2+}\left(g\right)+2e^{-}.....\left(b\right)$

$M\left(g\right)\to M^{+}\left(g\right)+e^{-}.....\left(c\right)$

$M^{+}\left(g\right)\to M^{2+}\left(g\right)+e^{-}.....\left(d\right)$

$M\left(g\right)\to M^{2+}\left(g\right)+2e^{-}.....\left(e\right)$

Energy required to change $M\left(s\right)$ into $M^{+}\left(g\right)$ is -

If the anions $\left(A\right)$ form hexagonal closed packing and cations $\left(C\right)$ occupy $2/3^{rd}$ of all octahedral voids in it, then the general formula of the compound is:

Which uses more energy, a 250 W TV set in 1 h, or a 1200 W toaster in 10 minutes?

Conjugate base of $HC{O}_3^{-}$ in the given is:

Reaction :$HC{O}_3^{-}+H_{2}O\to C{O}_3^{2-}+H_3{O}^{+}$

The critical temperature of water is higher than that of $O_2$ because the water molecule has

The pH of blood does not appreciably change by a small addition of an acid

or a base because blood

Which of the following solutions have the same concentration ?

(a) 20 g of NaOH in 200 mL of solutoin

(b) 0.5 mol of KCl in 200 mL of solution

(c) 40 g of NaOH in 100 mL of solution

(d) 20 g of KOH in 200 mL of solution

Predict the product from the given $E2$ reaction.

Electronic configuration of $\left[Cu\right(N{H}_3{)}_6{]}^{2+}$ on the basis of crystal field splitting theory is :

The decomposition of hydrogen peroxide in an aqueous solution is a reaction of first order. It can be followed by titrating 10 mL portions of reaction mixture at various times from the beginning of reaction against a standard solution of $KMn{O}_4$. Volume of $KMn{O}_4$ solution used in each case is proportional to the remaining concentration of $H_2{O}_2$. From the following data, calculate the rate constant of the reaction.

$\begin{array}{cccc}\text{Time (in seconds)}& 0& 600& 1200\\ KMn{O}_4\text{sol. used (in mL)}& 22.8& 13.8& 8.20\end{array}$

Which of the following is soluble in alcohol?

The electrophile involved in the above reaction is:

Which of the following species / molecules is aromatic?

The absorption transition between the first and the fourth energy states of hydrogen atom are $3$. The number of emission transitions between these states will be

The wave number of energy emitted when an electron jumps from $4^{th}$ orbit to $2^{nd}$ orbit in hydrogen is $20.497{\text{cm}}^{-1}$. The wave number of energy for the same transition in ${\text{He}}^{+}$ is-

Rate constant for a $n^{th}$ order reaction has the dimensions
(Where M is concentration unit)

The
HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint:
Can be explained on the basis of sp³
hybridisation in NH₃
and only s−p bonding
between hydrogen and other elements of the group].