Assuming the existence of the compound shown that is given trivial name, “Ravana” what would you call it in IUPAC?
Category Archives: Chemistry
The label on the bottle of $H_2O_2$ solution …
$N{H_2} – CO – N{H_2}\xrightarrow{NaOBr} {N_2}$
Hydrazine formed from Urea is further oxidised to $N_2$ as shown below:
$N{H_2} – \mathop {\mathop C\limits^\parallel }\limits^O – N{H_2} \xrightarrow[\text{Hoffmann Bromamide Degradation}]{\text{NaOBr}} N{H_2} – N{H_2} \longrightarrow {N_2}$
Assuming 1 dL of blood sample contains 30 mg of Urea, calculate the volume of $N_2$ gas obtained at NTP from the sample. Continue reading $N{H_2} – CO – N{H_2}\xrightarrow{NaOBr} {N_2}$
Nomenclature of Crown Ether
[18]-crown-6 has:
A) 18 Carbon & 6 Oxygen atoms
B) 6 Carbon & 18 Oxygen atoms
C) 12 Carbon & 6 Oxygen atoms
D) 6 Carbon & 12 Oxygen atoms
Key
Crown ether named [T]-crown-O, means T is the total number of atoms in the ring and O is the number of oxygen atoms in the ring.
For, [18]-crown-6 there are 6 Oxygen atoms and a total of 18 atoms.
So, number of Carbon atoms = 18 – 6 = 12
Hence, option (C).
The volume V of an enclosure contains a mixture ….
The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :
(A) $\frac {3RT}{V}$ (B) $\frac {5}{2} \frac {RT}{V}$
(C) $\frac {88RT}{V}$ (D) $\frac {4RT}{V}$ Continue reading The volume V of an enclosure contains a mixture ….
A 6.50 molal solution of KOH (aq) has density of $1.89 g.cm^{-3} $ ….
A 6.50 molal solution of KOH (aq) has density of $1.89 g.cm^{-3} $. The molarity of solution is _ _ _ _ $mol.dm^{-3} $. (Round off to the nearest integer).
[Atomic masses: K = 39.0 u, O = 16.0 u, H = 1.0 u] Continue reading A 6.50 molal solution of KOH (aq) has density of $1.89 g.cm^{-3} $ ….
For the reaction $A(g) \rightleftharpoons B(g)$ at 495 K ….
For the reaction $A(g) \rightleftharpoons B(g)$ at 495 K, $\Delta_r G^\circ = -9.478 kJ.mol^{-1} $. If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is _ _ _ _ millimoles. (Round off to the nearest integer)
[$R=8.314 J.mol^{-1} .K^{-1} $; ln 10 = 2.303 ] Continue reading For the reaction $A(g) \rightleftharpoons B(g)$ at 495 K ….