A metal M forms a volatile carbonyl complex $M(CO)_x$. 0.525g of this complex is heated in a closed vessel of volume 250 mL at $50^\circ C$. The pressure inside the vessel after complete decomposition of the complex is 746 Torr. If the gas constant R is 0.0821 L atm $K^{-1} mol^{-1}$, what is the value of x in the formula $M(CO)_x$? (Assume the metal M does not contribute significantly to the pressure)

The number of moles of $KMn{O_4}$ that will be needed to react completely with one mole of oxalate ion in acidic solution is

In the reaction, ${\rm{VO}} + {\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3} o {\rm{FeO}} + {{\rm{V}}_2}{{\rm{O}}_5}{\rm{.}}$. The eq.wt. of ${{\rm{V}}_2}{{\rm{O}}_5}$ is equal to its :

What weight of ${\rm{HN}}{{\rm{O}}_3}$ is needed to convert $5\,g\,{I_2}$  into ${\rm{HI}}{{\rm{O}}_3},{\rm{HN}}{{\rm{O}}_3} o {\rm{NO}}?$

What volume of 0.40 $M$ ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ would be required to react with the ${{\rm{I}}_2}$ liberated by adding excess of KI to 50 mL of 0.20 $M$ ${\rm{CuS}}{{\rm{O}}_4}$ solution?

The number of moles of ${K_2}C{r_2}{O_7}$ that will be needed to react with one mole of nitrite ion in acidic solution is

How many litre of ${\rm{C}}{{\rm{l}}_2}$ at STP will be liberated by the oxidation of ${\rm{NaCl}}$ with 10 g ${\rm{KMn}}{{\rm{O}}_4}$?

In the following change, $3{\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} o {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + 4{{\rm{H}}_2}$. If the atomic weight of iron is 56, then its equivalent weight will be :

What is the normality of a ${\rm{KMn}}{{\rm{O}}_4}$ solution to be used as an oxidant in acid medium, which contain 15.8 g of the compound in 100 mL of solution? Mol. wt. of ${\rm{KMn}}{{\rm{O}}_4}$ is 158 :

$0.2$ g of a sample of ${{\rm{H}}_2}{{\rm{O}}_2}$ required $10$ mL of $N{\rm{KMn}}{{\rm{O}}_4}$ in a titration in the presence of ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}{\rm{.}}$. Purity of ${{\rm{H}}_2}{{\rm{O}}_2}$ is :

In the following change, $3{\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} o {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + 4{{\rm{H}}_2}$. If the atomic weight of iron is 56, then its equivalent weight will be :