Tag: chemistry

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Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Which of the following forms a white precipitate when added to a solution of $NaCl$?

  1. $N _2$
  2. $KI$
  3. $CCl _4$
  4. $AgNO _3$
  5. $CaCO _3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Silver nitrate $\displaystyle AgNO _3$ forms a white precipitate of $AgCl$ when added to a solution of $NaCl$. 

$\displaystyle AgNO _3 + NaCl \rightarrow AgCl \uparrow + NaNO _3$

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

What is the key element in soil fertilizer?

  1. Carbon

  2. Nitrogen

  3. Oxygen

  4. Neon

  5. Argon
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Nitrogen is the key element in soil fertilizer. It is essential for plant growth but plants cannot utilize atmospheric nitrogen as it is inert due to presence of triple bond. Hence, it is supplied through fertilizer.

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Which of the following statements regarding nitrogen sesquioxide is not correct?

  1. Nitrogen sesquioxide is stable only in the liquid state. It dissociated in the vapour phase

  2. Dinitrogen sesquioxide is a neutral oxide
  3. Dinitrogen sesquioxide contains a weak $N - N$ bond

  4. Dinitrogen sesquioxide exists in two different crystalline forms
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Nitrogen sesquioxide or dinitrogen trioxide $(N _2O _3)$ is an acidic oxide. It is an anhydride of nitrous acid and dissolves in water to form $HNO _2$, an unstable acid.

$N _2O _3 + H _2O \rightarrow 2HNO _2$
Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Which of the following is the density of nitrogen at STP?

  1. $0.33g/L$
  2. $0.65g/L$
  3. $0.80g/L$
  4. $1.25g/L$
  5. $1.60g/L$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At STP, one mole of any gas occupies 22.4litres of volume. Nitrogen exists in molecular form as $N _2$. 

So, $Denisty= \dfrac{\text Molecular \ mass}{\text Volume}$

Here a Molecular mass of one mole of $N _2$. is 14(2)=28 grams and Volume is 22.4 litres.
So $Denisty= \dfrac{28}{22.4}g/L = 1.25g/L$.
Option D is correct

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Which of the following gas, when ignited, burns with a blue flame and is not very soluble in water?

  1. $O _2$
  2. $CO _2$
  3. $N _2$
  4. $He$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Nitrogen when ignited in presence of oxygen it forms various oxides of nitrogen like $NO,\ NO _2$ and burns with a blue flame.
Hence, option $C$ is correct.
Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

${ N } _{ 2 }(g)\ +\ { 3H } _{ 2 }(g)\rightleftharpoons { 2NH } _{ 3 }(g)$


For the reaction initially the mole was 1 : 3 of $N _2$ and ${ H } _{ 2 }$. At equilibrium 50% of each has reacted. If the equilibrium pressure is p, the partial pressure of ${ NH } _{ 3 }$ at equilibrium is :

  1. $\dfrac { p }{ 3 }$
  2. $\dfrac { p }{ 4 }$
  3. $\dfrac { p }{ 6 }$
  4. $\dfrac { p }{ 8 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$N _2 + 3H _2⇌2NH _3$
  1           3             0
(1-$x$)  ($3-3x$)     $2x$
            
$1-x =  \dfrac{1}{2}$
$\therefore x=\dfrac{1}{2}$
Mole of $H _2 = $ 1.5 mol
Mole of $N _2 = $ 0.5 mol
Mole of $NH _3 = $ 1.0 mol
Now,
Mole fraction of $NH _3$ =$\dfrac{1}{3}$

partial pressure = $p*\dfrac{1}{3}=\dfrac{p}{3}$
Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Gas B is paramagnetic and support the combustion compound B is :

  1. $N _2$
  2. $N _2O$
  3. $O _2$
  4. $Na _2O$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$O _2$ is paramagnetic due to the presence of $2$ unpaired $e^-$ in the $\Pi$-antibonding orbital.

$\sigma1s^2\sigma^*1s^2\sigma2s^2\sigma^*2s^2\sigma2p _3^2\Pi2p _x^2=\Pi2p _y^2\Pi^*2p _x^1=\Pi^*2p _y^1$
Also, $O _2$ supports combustion as it does not burn but it is an oxidizer, it oxidizes other reactive materials, itself getting reduced.