VITEEE 2017 :: Chemistry :: General questions

• Chemistry - General questions

Which group contains coloured ions out of

1. $Cu^{2+}$     2. $Ti^{4+}$

3. $Co^{2+}$     4. $Fe^{2+}$

Answer:

Explanation:

$Cu^{2+}[Ar]3d^{9}, Ti^{4+}[Ar]3d^{0} Co^{2+}[Ar]3d^{7},Fe^{2+}[Ar]3d^{6}$

1,3,4 are coloured ions hence the answer is b.

1 -Propanol and 2-propanol can be distinguished by

Answer:

Explanation:

(a) $CH_{3}CH_{2}CH_{2}OH\xrightarrow[{or K_{2}Cr_{2}O_{7}/H^{+}}]{alk.KMnO_{4}}CH_{3}CH_{2}CHO$

$\rightarrow CH_{3}CH_{2}COOH\xrightarrow{Fehling sol.} No Reaction$

(b) $CH_{3}CHOHCH_{3}\xrightarrow[{or K_{2}Cr_{2}O_{7}/H^{+}}]{alk.KMnO_{4}}CH_{3}COCH_{3}$

$\rightarrow CH_{3}COOH\xrightarrow{Fehling sol.} No Reaction$

(c) $CH_{3}CH_{2}CH_{2}OH\xrightarrow{Cu}CH_{3}CH_{2}CHO\xrightarrow{Fehling sol.}Cu_{2}O_{(Red)}$

$CH_{3}CHOHCH_{3}\xrightarrow{Cu}CH_{3}COCH_{3}\xrightarrow{Fehling sol.}No reaction$

(d) $CH_{3}CH_{2}CH_{2}OH$ OR

$CH_{3}CHOHCH_{3}\xrightarrow{Conc.H_{2}SO_{4}}CH_{3}CH=CH_{2}$

For a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow the equation

log k = - (2000)1/T + 6.0 . The pre-exponential factor A and the activation energy E_a respectively, are

Answer:

Explanation:

$\log_{}{k}=\log_{}{A}-\frac{E_{a}}{2.303RT}$   ....(1)

Also given log k = 6.0 - (2000)1/T    .....(2)

On comparing equations, (1) and (2)

log A = 6.0 → A = 10⁶ s^-1

and $\frac{E_{a}}{2.303R}$ = 2000

→ E_a = 2000x2.303 x 8.314 = 38.29kJmol^-1

If 'a' stands for the edge length of the cubic systems : simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively

Answer:

Explanation:

Following generalization can be easily derived for various types of lattice arrangements in cubic cells between the edge length (c) of the cell and r the radius of the sphere

For simple cubic: a = 2r or r = a/2

For body centred cubic : a = $\frac{4}{\sqrt{3}}$ r or r = $\frac{\sqrt{3}}{4}$a

For face centred cubic : a = $2\sqrt{2}r$ or r = $\frac{1}{2\sqrt{2}}a$

Thus the ratio of radii of spheres for these will be

simpte : bcc : fcc = $\frac{1}{2}a:\frac{\sqrt{3}}{4}a:\frac{1}{2\sqrt{2}}a$

option (a) is correct answer.

Saponification of coconut oil yields glycerol and

Answer:

Explanation:

Saponification (alkaline hydrolysis) of oils and fats gives glycerol and sodium salt of  fatty acids, which is sodium palmitate in the present question

• Chemistry - General questions