VITEEE 2015 :: Chemistry :: General questions

• Chemistry - General questions

The molarity of NO₃ in the solution after 2L of 3 M AgNO₃ is mixed  with 3L of 1M BaCl₂ is 

Answer:

Explanation:

 2L of 3M AgNO₃ will contain 6 moles of  AgNO₃.

3L of 1M  BaCl₂ will contain 3 moles of BaCl₂

$2AgNO_{3}+BaCl_{2}\rightarrow 2AgCl+Ba(NO_{3})_{2}$

 So, 6 moles of AgNO₃ will react with 3 moles on BaCl₂ it means , two solution will react completely to form 3 moles of 

 Ba(NO₃)₂ =6 moles of NO₃^-  ions in 2⁺ 

3=5 L solution

Hence, molarity of  $NO_{3}^{-}$=6/5=1.2 M

For the given reaction,

$H_{2}(g)+Cl_{2}{}(g)\rightarrow 2H^{+}(aq)+2Cl^{-}(aq).$

 $\triangle G^{0}=-262.4kJ$ 

 The value of free energy of formation

( $\triangle G^{0}_{f}$ ) for the ion Cl^-1 (aq)  . therefore will be

Answer:

Explanation:

$(\triangle G^{0}) reaction =\triangle G^{0}_{f}$   (products ) - $\triangle G^{0}_{f}$ (reactants)

$\therefore$    $264.4=[2\triangle G^{0}_{f} (H^{+})+2 \triangle G^{0}_{f}(C^{1-})]$

or   $264.4=-[\triangle G^{0}_{f} (H_{2})+ \triangle G^{0}_{f}(Cl_{2})]$

 $=[0+2\triangle G^{0}_{f}  (Cl^{-})]+[0+0]$

 or   $-262.4=2\triangle G^{0}_{f}(Cl^{-})$

 or  $-\triangle G^{0}_{f}(Cl^{-})=-131.2 kJmol^{-1}$

$XCl_{2}(excess)+ YCl_{2}\rightarrow XCl_{4}+Y\downarrow$

$YO$    $\underrightarrow{\triangle}$    $\frac{1}{2}O_{2}+Y$

          >400°C

 Ore of Y would be

Answer:

Explanation:

$SnCl_{2}+HgCl_{2}\rightarrow SnCl_{4}+Hg$

   $(xCl_{2})$  $(yCl_{2})$    $(xCl_{2})$   (y)

 $HgO$    $\underrightarrow{\triangle}$   $Hg+ \frac{1}{2}O_{2}$

                >400° c   (y) 

 So ore of  $\gamma$ is HgS , i.e, cinnabar

The chemical composition of slag formed during the smelting process in the extraction of copper is 

Answer:

Explanation:

During smelting process of copper from copper pyrites reaction are 

$Cu_{2}O+FeS\rightarrow Cu_{2}S+FeO$

$2FeS+3O_{2}\rightarrow2FeO+2SO_{2}$

Consider the water gas equilibrium reaction, 

$C(s)+H_{2}O(g)\rightleftharpoons CO(g)+H_{2}(g)$

 which of the following statement is true at equilibrium?

Answer:

Explanation:

$K=\frac{[CO(g)][H_{2}(g)]}{[H_{2}O(g)]}$

Concentration will increase, on halving the volume. There are two terms in numerator. So to keep K constant, the concentration of [H₂ O] should increase much more.

• Chemistry - General questions