An unbiased coin is tossed to get $2$ points for turning up a head and $1$ point for the tail. If three unbiased coins are tossed simultaneously,then the probability of getting a total of an odd number of points is

$A$ compound $X$ upon reaction with $H_{2}O$ produces a colorless gas $Y$ with a rotten fish smell. Gas $Y$ is absorbed in a solution of $CuSO_{4}$ to give $Cu_{3}P_{2}$ as one of the products. Predict the compound $X$.

$A$ volume of $x \ mL$ of $5 \ M \ NaHCO_3$ solution was mixed with $10 \ mL$ of $2 \ M \ H_2CO_3$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of $235.3 \ mV$, then the value of $x = . . . . . . \ mL$ (nearest integer).
$Sn_{(s)} \mid Sn(OH)_6^{2-}(0.5 \ M) \mid HSnO_2^{-}(0.05 \ M) \mid OH^{-} \mid Bi_2O_{3(s)} \mid Bi_{(s)}$
Consider upto one place of decimal for intermediate calculations.

Given :	$E^{\circ}_{HSnO_2^{-} \mid Sn(OH)_6^{2-}} = -0.9 \ V$
	$E^{\circ}_{Bi_2O_3 \mid Bi} = -0.44 \ V$
	$pKa_{(H_2CO_3)} = 6.11$
	$\frac{2.303 \ RT}{F} = 0.059 \ V$
	$Antilog(1.29) = 19.5$

$A$ solution containing $10 \ g$ per $dm^3$ of urea (molecular mass $= 60 \ g \ mol^{-1}$) is isotonic with a $5 \%$ solution of a non-volatile solute. The molecular mass of the non-volatile solute is ........ $g \ mol^{-1}$.

If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to

During meiosis $I,$ the chromosomes start pairing at