TS EAMCET 2018 :: Chemistry :: General questions

• Chemistry - General questions

 In reaction  (1), $XeF_{6}$  hydrolysis  to form HF and X . In reaction (2), $XeF_{6}$  on partial hydrolysis form HF, Y and Z

 The products X, Y, Z respectively, are 

Answer:

Explanation:

 Complete hydrolysis of $XeF_{6}$ produces xenon trioxide

     $XeF_{6}+3H_{2}O \rightarrow   XeO_{3}+6HF$

whereas partial hydrolysis  of $XeF_{6}$  form different  products

   $XeF_{6}+2H_{2}O \rightarrow   XeO_{2}F_{2}+4HF$

  $XeF_{6}+H_{2}O \rightarrow   XeOF_{4}+2HF$

 In the Freundlich  isotherm  $\left(\frac{x}{m}=k_{p}^{1/n}\right)$ plot of $\log \frac{x}{m}$ vs $\log p$, the intercepts are (where,x,m, p and k are mass of the gas, mass of adsorbent, pressure and constant  which depend on the nature of adsorbent, respectively)

Answer:

Explanation:

 For Freundlich adsorption isotherm equation

   $\frac{x}{m}=k_{p}  \frac{1}{n}$

 or     $\log \frac{x}{m}= \log k+  \frac{1}{n} \log p$

 A plot  of $\log \frac{x}{m}$ versus $\log p$  is linear  with slope = $\frac{1}{n}$  and intercept = $\log k$

$PCl_{5}\rightleftharpoons PCl_{3}+PCl_{2}$

 if the equilibrium constant $(K_{C})$  for the above reaction at 500 K is 1.79 and the equilibrium concentration of $PCl_{5}$  and $PCl_{3}$  are 1.41 M  and 1.59 M . respectively , then the concentration of $Cl_{2}$  is approximately 

Answer:

Explanation:

 Equilibrium constant $ (K_{C})=  \frac{ [PCl_{3}][Cl_{2}]}{[PCl_{5}]}$

 $1.79= \frac{1.59 \times [Cl_{2}]}{1.41}$

 $[Cl_{2}]= \frac{1.79 \times 1.41}{1.59}= 1.5877 $ ≈ 1.59 m

 Kinetic energy in kJ of 280 g of $N_{2}$ at $27^{0} C$ is approximately (R=8.314 J mol^-1K^-1)

Answer:

Explanation:

280g  of $N_{2}$=$\frac{280}{28}=10 $ moles of $N_{2}$

 $27^{0} C= (27+273)K=300K$

 K.E =   $  \frac{3nRT}{2}$ =  3 x number of mol x 3 x moles of gas x gas constant x temperature /2

  = $\frac{3 \times 10 \times 8.314 \times 300}{2}=37.4 \times 10^{3} J$=37.4  kJ

Which of the following set of radiations  can not be seen in hydrogen atomic spectrum ?

(i)  $\gamma-$ radiation   (ii) Uv  (iii) X-rays   (iv)  Infrared

Answer:

Explanation:

 In hydrogen atomic spectra the five series have frequency range as:

 Lyman seris $\rightarrow$  UV region 

 Balmer series $\rightarrow$ Visible region

Paschen series  $\rightarrow$ IR region

Brackett series $\rightarrow$ For I R region

Pfund series $\rightarrow$   IR region

• Chemistry - General questions