Tag: chemistry

Quizzes (490) Questions (9892)

Questions Related to chemistry

Multiple choice chemistry adsorption and colloids applications of colloids colloids around us the colloidal state

Silver iodide is used for producing artificial rains because $AgI$:

  1. is easy to spray at high altitude

  2. is insoluble in water

  3. is easy to synthesize

  4. has crystals similar to ice
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Solid silver iodide is preferred as a nucleator for artificially stimulated rain because its crystal lattice parameters are close to those of ice, and for that reason are more effective than other salts in causing small ice crystals in clouds to melt.

Multiple choice chemistry adsorption and colloids applications of colloids colloids around us the colloidal state

To stop bleeding from an injury ferric chloride can be applied
Which comment about the statement is justified?

  1. It is not true, ferric chloride is a poison
  2. It is true, ${Fe}^{3+}$ ions coagulate blood which is negatively charged sol
  3. It is not true, ${Cl}^{-}$ ions from positively charged sol, profuse bleeding takes place
  4. It is true, coagulation takes place because of formation of negatively charged sol with ${Cl}^{-}$ ions.

  5. It is not true, ferric chloride is ionic and gets into blood stream
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To stop bleeding from an injury, Ferric chloride can be applied because of the following reason:

(A) $Fe^{3+}$ ions coagulate blood which is a negatively charged solution
(B) $Cl^−$ ions coagulate blood which is a positively charged solution

Blood being a colloidal solution, its coagulation by Hardy-Shuze's law states " Higher the charge on cation higher will be its efficiency to coagulate the colloidal solution" . Thus Option B is correct.

Multiple choice chemistry structure of the atom electronic configuration and valency electron configuration periodic trends in physical properties

An atom of each element has a definite combining capacity called : 

  1. valency

  2. affinity

  3. bonding

  4. energy levels
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An atom of each element has a definite combining capacity called its valency.

For example Na has electronic configuration 2,8,1 
So, valency is 1 as after losing one electron it will have stable octet.

Multiple choice chemistry structure of the atom electronic configuration and valency electron configuration periodic trends in physical properties

The valency of nitrogen in nitrogen dioxide is:

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(D)$  $ 4$


$Sol. $  let the valency of nitrogen in be $ x$. 
           valency of oxygen = $ -2 $
          For $NO _2 $ it will be ,
                                $ x + 2 \times(-2) = 0 $
                                 $ x = +4 $
Hence, the valency of nitrogen in $ NO _2$ is $ +4 $. 

Multiple choice chemistry structure of the atom electronic configuration and valency electron configuration periodic trends in physical properties

Valency of magnesium and oxygen in $MgO$ is:

  1. one and one

  2. two and two

  3. one and two

  4. two and one
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ans. $(B)$ $(two$ $and$ $two)$

This is because magnesium loses two electrons to have an octet and oxygen gains two electrons to have an octet. The final formula of magnesium oxide is ${MgO}$. So, the valencies magnesium and oxygen in $ {MgO}$ are $2$ and $2$ .
Multiple choice chemistry periodic table electronic configuration and valency electron configuration periodic trends in physical properties

The ionisation potential of hydrogen atom is $13.6\ eV$. The energy of required to remove an electrons in the $n=2$ state of hydrogen atom is:

  1. $27.2\ eV$
  2. $13.6\ eV$
  3. $6.8\ eV$
  4. $3.4\ eV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$E _n=-13.6\ ev\left (\dfrac {Z^2}{n^2}\right)=\dfrac {-E _1}{n^2}[I.E _1 =-E _1 =13.6\ eV \Rightarrow E _1=-13.6\ eV]$

$\therefore E _2 =\dfrac {-13.6\ ev}{4}=-3.4ev$

$\therefore I.E _2=-E _2 =3.4\ ev$.

Option D is correct.