BITSAT Chemistry Test - 1 - JEE MCQ

Water gas and H₂ reacts to produce methanol.

1 s² 2s² 2p³ can be shown as 

Therefore, there are 3 unpaired electrons.

Primary amine on heating with CS₂ is presence of excess of HgCI₂ produce ethyl isothiocyanate which has smell of mustard oil. Therefore, it is known as Hoffmann’s mustard oil reaction.

The most stable carbonium ion is triphenyl methyl carbonium ion because the π-electrons of three benzene rings are delocalized with tb vacant π-orbital of central carbon atom. Therefore, it is stabilized by resonance.

The electronic configuration of bromine atom is
1s², 2s² 2p⁶, 3s² 3p⁶ 3d¹⁰,4s² 4p⁵
Hence, total number of p-electrons is 17.

Flux is used to remove silica and undesirable metal oxide. It is an external material added during smelting to convert infusible impurities into an easily fusible material known as slag.

Since, acetic acid is a weak electrolyte, hence

Acetaldehyde and acetone both comes under carbonyl group. Both respond to 2, 4-dinitrophenyl hydrazine to produce yellow or orange precipitate. Aldehydes produce colour with Schiff’s reagent while ketones do not respond to this test.

On heating the one end of the metal piece the other end becomes hot this is due to the movement of energised electrons from one end to other end.

Rate = k[H₂O₂] [l−]
Order of reaction =1 + 1 = 2
Dimensions of 'k'= 

The given reaction started initially with [A] =2.00 M
Time taken, (t)=200 min and  [A] becomes 0.15 M
For a first order reaction, the rate constant is

(Where, a, a−x are the initial and final concentrations, respectively).

Further,

The Arrhenius equation gives the relations between rate constants at different temperatures as:

For the 1st reaction,


For the 2nd reaction,


or , 
Since the rate of disappearance of N2O5 is same in both reactions,
therefore, equation (i) = equation (ii)

Hence,  We have, k=2k'.

As the half life of C¹⁴ is 5760 years, so a 6 year old fossil’s age can’t be determined.
Further this technique cannot be used to date objects older than 30,000 
years.
After this length of time the radioactivity is too low to be measured.

C_t=C₀e^−Kt
According to the question,



Alternative solution :

When [A] = [B]

At 2^oC (275 K) , the reaction is three times slower than at 27^oC (300 K) . This implies that for souring of milk.

Applying Arrhenius equation 

Therefore, E_a=30.146 (kJ mol⁻¹).

Let the order with respect to A and B is x and y respectively.
Hence,
Rate r=[A]^x[B]^y ...(i)
On doubling the concentration of A, rate increases 4 times,
4r=[2A]^x[B]^y ... (ii)
From Eqs. (i) and (ii)

∴  x = 2
∴ Order with respect to A is 2
If concentration of A and B both are doubled,
8r=[2A]^x[2B]^y  ... (iii)
From Eqs. (i) and (iii), we get

2^y=2
∴ y = 1
Hence , differential rate equation is

[Where, C_A and C_B = concentrations of A and B]

Mass no. = no. of protons + no. of neutrons.

ψ has no physical significance while ψ2 represnts the probability density of finding an electron.

Explanation

• Anomalous Zeeman Effect: The spin quantum number was introduced to account for the anomalous Zeeman effect, which is the splitting of emission lines into more lines than predicted by the magnetic quantum number. This effect occurs when the electron spin interacts with an external magnetic field, causing the energy levels to split further based on the spin states of the electron.


• Representation of Spin States: The two spin states of the electron, represented by the arrows ↑ (spin up) and ↓ (spin down), help differentiate between the two possible orientations of the electron's intrinsic angular momentum. This distinction is crucial in understanding how the electron interacts with external magnetic fields and contributes to the splitting of emission lines.


• Quantum Mechanical Explanation: In quantum mechanics, the spin quantum number is a fundamental property of particles like electrons, and it plays a significant role in determining the behavior of these particles in different physical situations. The introduction of the spin quantum number allows for a more comprehensive description of electron states and their interactions with external fields.


 

This is 1-phenyl propan-1-one.

Reductive Ozonolysis of alkene forms aldehyde or ketone.

C₂H₅OH get oxidizes to CH₃CHO and –COCH₃ group is important for iodoform.

Because conjugate base that form will be stable in case of Ketone.

This is a case of dilution.

Thus, 3.846 L of 15.6 M solution is diluted to 10.0 L by addition of 6.154 L of H₂0.

125 mL of 0.05 M AgN0₃ = 125 x 0.05 millimoles