MCQ
Nitrolim is a:
  • A
    Mixture of calcium carbide and nitrogen
  • Mixture of calcium cyanamide and carbon
  • C
    Mixture of calcium cyanide and carbon
  • D
    Mixture of $\ce{NH_4​CN}$ and $\ce{CaCN}$

Answer

Correct option: B.
Mixture of calcium cyanamide and carbon
When nitrogen is passed over calcium carbide at a suitable high temperature, we get a mixture called nitrolim which is a nitrogenous fertilizer, which is a mixture of calcium cyanamide and carbon.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Which of the following compound is not resonance stabilized ?
The Lassaigne’s extract is boiled with dil. $HNO_3$ before testing for halogens because
Which one of the following configuration represents a noble gas
A solution of $\mathrm{H}_2 \mathrm{SO}_4$ is $31.4 \% \mathrm{H}_2 \mathrm{SO}_4$ by mass and has a density of $1.25 \mathrm{~g} / \mathrm{mL}$. The molarity of the $\mathrm{H}_2 \mathrm{SO}_4$ solution is _____________$\quad \mathrm{M}$ (nearest integer)[Given molar mass of $\mathrm{H}_2 \mathrm{SO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
$B$ has a smaller first ionization enthalpy than $Be$. Consider the following statements :

$(I)$ It is easier to remove $2 \mathrm{p}$ electron than $2 \mathrm{s}$ electron

$(II)$ $2 \mathrm{p}$ electron of $\mathrm{B}$ is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $Be.$

$(III)\; 2 s$ electron has more penetration power than $2 \mathrm{p}$ electron.

$(IV)$ atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$

(Atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )

The correct statements are

The synthesis of $3-$octyne is achieved by subsequent stepwise reactions of sodium amide with an alkyne, and a bromoalkane. The bromoalkane and the other alkyne respectively are:
Which of the following does not exists as ionic substance in solid state
Calculate change in internal energy if $\Delta H = -92.2\, kJ, P = 40\, atm$ and $\Delta H = -1\,L.$ ......$kJ$
When $\alpha $-particles are sent through a thin metal foil, most of them go straight through the foil because (one or more are correct)
Hydrogen bonding is $NOT$ responsible for