What is meant by diazonium salts?

What is meant by diazonium salts?

Solution-
Diazonium salts are the compounds that have the common functional group Ar/R-N2^(+).
the reaction to make the diazonium salts is as following-

It can be made by alkyl or aryl-amine in the presence of sodium nitrate and HCl,
Diazonium salts are used to make cyanide, fluoride, iodide and most importantly phenols. It is very easy to form like these compounds with diazonium salts otherwise the reaction to make the following compounds are very tough and time-consuming.

Hope this helps, keep looking this blog, I regularly upload useful content for readers.

Read More

Glucose can be tested by??

Glucose can be tested by ______ reagent??
Tollens Test:-
Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a ketone.
It exploits the fact that aldehydes are readily oxidized (see oxidation), whereas ketones are not.
Tollens’ test uses a reagent known as Tollens’ reagent, which is a colorless, basic, aqueous solution containing silver ions coordinated to ammonia [Ag(NH_3)^{2+}].

Benedict Test:-
Benedict's Test is used to test for simple carbohydrates.
The Benedict's test identifies reducing sugars (monosaccharide's and some disaccharides), which have free ketone or aldehyde functional groups.
Benedict's solution can be used to test for the presence of glucose in urine.

Read More

The helical structure of protein is stabilized by??

The helical structure of protein is stabilized by:
[A]hydrogen bonds
[B]ether bonds
[C]peptide bonds
[D]dipeptide bonds

Solution=

• The helix is common in the secondary structure of proteins and is a right hand-spiral conformation (i.e. helix) 
• In helix structure, every backbone N−H$N-H$ group donates a hydrogen bond to the backbone C=O$C=O$ group of the amino acid located three or four residues earlier along the protein sequence.

Read More

Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume. of the two gases (dry and at STP) produced will be approximately: (in litres) (A) 22.4 (B) 44.8 (C) 67.2 (D) 11.2

 Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume. of the two gases (dry and at STP) produced will be approximately: (in litres)
(A) 22.4
(B) 44.8
(C) 67.2
(D) 11.2

Solution=
According to the first law of Faraday,
Wt/Ewt=Q/F
QF= faraday No.
∴ charge of 1 mol of electrons is equal to 1 faraday.
Wt/Ewt=ne/F∴Q=ne and n=4
mole×V.f.=4F/F
mole×2=4
mols of H2O=2
Hence the two moles of H2O will be decomposed into its gaseous components
2H2O⟶2H2+O2
2moles of H2(at the cathode) and 1mol of O2(at the anode) will produce.
moles=v/22.4
So volume of H2 and O2 will be = 44.8+22.4 = 67.2

Read More

Calculate the volume of CO2 evolved by the combustion of 50 ml of a mixture contains C2H4 and 60% CH4 (by volume)

Calculate the volume of CO2 evolved by the combustion of 50 ml of a mixture contains C2H4 and 60% CH4 (by volume)

Solution=
50 ml mixture of C2H4 and CH4,

where CH4=50% of 60ml=30ml and C2H4=20ml.

applying limiting reagent law

C2H4 +CH4+5O2⟶3CO2+4H2O

(20−x) (30−x)3x
limiting reagent is C2H4

∴(20−x)=0 and x=20
putting the value of x for CO2
The volume of CO2=3x=3×20=60ml

Read More

A reaction follows the given concentration (M)− time graph. The rate for this reaction at 20 seconds will be: (A) 4 X 10^(-3) (B) 6 X 10^(-3) (C) 7 X 10^(-3) (D) 8 X 10^(-2)

A reaction follows the given concentration (M)− time graph. The rate for this reaction at 20 seconds will be:
(A) 4 X 10^(-3)
(B) 6 X 10^(-3)
(C) 7 X 10^(-3)
(D) 8 X 10^(-2)

Solution=
Draw a slope for 20 seconds
slope =−dy/dx=−dc/dt=−(0.3−0)/(40−0)  ≈7×10−3

Read More

Radiation corresponding to the transition n=4 to n=2 in hydrogen atoms falls on a certain metal (work function =2.5eV). The maximum kinetic energy of the photo-electrons will be: (A) .55ev (B) 2.55ev (C) 4.5ev (D) None

Radiation corresponding to the transition n=4 to n=2 in hydrogen atoms falls on a certain metal (work function =2.5eV). The maximum kinetic energy of the photo-electrons will be:
(A) .55ev
(B) 2.55ev
(C) 4.5ev
(D) None


Radiation corresponding to the transition
1/λ=RZ^2[1/n1^2−1/n2^2]

1/λ=RZ^2[1/2^2−1/4^2]

1/λ=RZ^2[3\16]

R=109678cm−1 and Z= Atomic no. of atom

1/λ=109678×12×316=20564cm^−1

λ=120564cm

KEmax=hc/λ−Φ

After putting the value of \lambda, h and c in CGS

KEmax= hc/λ−2.5ev=0.55ev

Read More