TS EAMCET 2019 :: Chemistry :: General questions

• Chemistry - General questions

 While combusting in air, 4g of $H_{2}$  gas was completely converted into water .If 36 $\mu$ mole of $CO_{2}$ from air is dissolved into that water, what is the concentration of $CO_{2}$?

Answer:

Explanation:

The equation takes place as follows;

 $2H_{2}+O_{2} \rightarrow   2H_{2}O$

  4g                            36g

 $\because$  36 g of $CO_{2}$ is dissolved  in 36 g of $H_{2}O$

$\therefore$   Conc. of $CO_{2}$= $\frac{36 \times 10^{-6} \times 10^{3}}{36}$

                                           $\left(\because conc:(C)=\frac{n}{v}\right)$

 = $10^{-3}$ M=1mM

 Hence , option (b) is the correct answer.

The variation of compressibility  factor (Z) with pressure (p in bar)  for some gases are shown in the figure below, Identify  the gases (A),(B) and (C) respectively

Answer:

Explanation:

The extent of deviation shown by real gases from ideal behaviour is measured in terms of compressibility factor (Z)  which is defined as the ratio of pV and nRT.

Mathematically,

              $Z=\frac{pV}{nRT}$

for ideal gases, Z=1

 Easily liquefiable gases such as $CO_{2}$ exhibit larger deviation as compared to gases such as $H_{2}, O_{2},N_{2}$ etc., that liquefy with difficulty.

Diffusion of $CH_{4}(g) $ and $O_{2}$  (g) occurs under similar conditions, then the ratio of their rates of diffusion is 

Answer:

Explanation:

From Grahjam 's law of diffusion:

$\frac{r_{CH_{4}}}{r_{O_{2}}}=\sqrt{\frac{M_{O_{2}}}{M_{CH_{4}}}}$

here, $r_{CH_{4}}$= rate of diffusion of $ CH_{4}$ gas

      $r_{O_{2}}$= rate of diffusion of $O_{2}$ gas

$M_{CH_{4}}$= molecular mass of $CH_{4}$=16

and   $M_{O_{2}}$ = molecular mass of $O_{2}$ gas=32

Hence,   $\frac{r_{CH_{4}}}{r_{O_{2}}}=\sqrt{\frac{32}{16}}=\sqrt{2}=1.414$

How many ions of the following have bond order of 2.5?

  $N_{2}^{-},NO^{-}, C_{2}^{-}, N_{2}^{+}, C_{2}^{2-}, CN^{+}$

Answer:

Explanation:

 bond order= (Number of bonding electrons- Number of anti-bonding electrons)/ 2

 Bond order of $N_{2}^{-}$ is $\frac{(10-5)}{2}= \frac{5}{2}$=2.5

Bond order of $NO^{-}$ is $\frac{(10-6)}{2}=\frac{4}{2}=2$

Bond order  of $C_{2}^{-}$   is $\frac{1}{2} (9-4)=\frac{5}{2}$=2.5

Bond order of $N_{2}^{+}$ is $\frac{(9-4)}{2}=\frac{5}{2}$=2.5

Bond order of $C_{2}^{2-}$ is $\frac{(10-4)}{2}=\frac{6}{2}=3$

 Bond order of $CN^{+}$ is $\frac{(8-4)}{2}=2$

The radius of the $2^{nd}$  orbit of $B^{4+}$ ion is 

Answer:

Explanation:

Radius of the nth orbit (r_n)= $0.529 \times \frac{n^{2}}{Z}$  Å

 For $B^{4+}$ion, (Z) =5, second  orbit (n)=2

 $r_{2}(B^{4+})= \frac{0.529 \times (2)^{2}}{5}$ Å= $\frac{0.529 \times 4}{5}$Å

 $r_{2}(B^{4+})$= 0.4232 Å

• Chemistry - General questions