The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ gives:

$A$ body of mass $2 \, kg$ slides down a curved track which is a quadrant of a circle of radius $1 \, m$. All the surfaces are frictionless. If the body starts from rest,its speed at the bottom of the track is ................. $m/s$.

The electrons,identified by quantum numbers $n$ and $l,$ $(i)$ $n = 4, l = 1$ $(ii)$ $n = 4, l = 0$ $(iii)$ $n = 3, l = 2$ $(iv)$ $n = 3, l = 1$ can be placed in order of increasing energy,from the lowest to highest,as

When acetaldehyde is reacted with $LiAlH_4$,what is the product formed?

The inverse of the matrix $\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ is

The required amount of $KBr$ (molar mass $= 119 \, g/mol$) in grams to start the precipitation of $AgBr$ in $500 \, mL$ solution of $0.05 \, M \, AgNO_3$ will be ($K_{sp}$ of $AgBr = 5 \times 10^{-13}$)