MHT CET 2020 :: Chemistry :: General questions

• Chemistry - General questions

 Identify the correct decreasing  order of densities of d-block elements

Answer:

Explanation:

 The  correct order of densities of d- block elements are as follows:

 Ni > Fe > Cr > V

  On moving across a period electrons have the same principal quantum number and so they are not making the atom bigger, but at the same time there is more protons and neutrons resulting increase in the number of nuclei and due to an effective increase in charge of the nucleus there is decreases in a radius of the atom.

$density =\frac{mass}{volume}$

 mass is increasing and volume is decreasing, overall we conclude that on moving left to right density increases

 For the reaction ,

 $N_{2}(g)+3H_{2}\rightarrow2NH_{3}(g) ; \triangle H$ is equal to

Answer:

Explanation:

 For the reaction,

$N_{2}(g)+3H_{2}$ $\rightarrow$ $2NH_{3}(g) $

 $\triangle$ H is equal to $\triangle$ U=-2RT

$\because$      $\triangle H= \triangle U+ \triangle  nRT $

 For the given reaction,

$\triangle$ n , is the difference between the number of moles of product and the number of moles of reactants.

 $\triangle$ n= (2-4) =-2

 Hence, $\triangle H= \triangle U-2RT $ 

 A first-order reaction is 25 % completed in 40 minutes. What is the rate constant k for the reaction?

Answer:

Explanation:

 Given,

 Time (t)= 40 min

 a=100

 (a-x)=100-25=75

 First order  reaction,

 $Rate(k)=\frac{2.303}{t}\log\left(\frac{a}{a-x}\right)=\frac{2.303\times\log 1.33}{40}$

 Which carbon atoms of $\alpha$- D- glucopyranose and $\beta $- D- fructofuranose respectively are linked together to form glycosidic linkage  in sucrose ?

Answer:

Explanation:

C-1 of $\alpha$-glucopyranose and C-2 of $\beta $- D fructofuranose respectively are linkage together to form glycosidic linkage in sucrose. It is shown  below:

 An element has a bcc structure with cell edge of 288 pm . The density of element is 7.2 g cm^-3. what is the atomic mass of an element ?

Answer:

Explanation:

 Given ,

 Cell in bcc, so Z=2

 Edge length (a) =288 pm

                        =288 X 10^-8 cm

 Density of metal (d) = 7.2 g cm^-3

                  N_A= 6.022 x 10²³

 We know that , density (d)= $\frac{Z \times M}{a^{3}.N_{A}}$

$\therefore M=\frac{d.a^{3}.N_{A} }{Z}$

$=\frac{7.2\times (2.88\times 10^{-8})\times6.022\times 10^{23}}{2}$

$=\frac{7.2\times 23.8878\times 10^{-24}\times6.022\times 10^{23}}{2}$

= 51.78 g mol^-1

• Chemistry - General questions