The correct answer is (C) : 53.33
\(n _{ c }= n _{ co _{2}}=\frac{2.64}{44}=0.06\)
\(n _{ H }=2 \times n _{ H _{2} O }=\frac{1.08}{18} \times 2=0.12\)
\(m _{0}=1.80-12 \times \frac{2.64}{44}-\frac{1.08}{18} \times 2\)
\(\quad=1.80-0.72-0.12=0.96 gm\)
\(\% 0=\frac{0.96}{1.80} \times 100=53.33 \%\)
List-I | List-II | ||
(A) | [Co(NH3)5(NO2)]Cl2 | (I) | Solvate isomerism |
(B) | [Co(NH3)5(SO4)]Br | (II) | Linkage isomerism |
(C) | [Co(NH3)6] [Cr(CN)6] | (III) | Ionization isomerism |
(D) | [Co(H2O)6]Cl3 | (IV) | Coordination isomerism |
List-I | List-II | ||
(A) | 1 mol of H2O to O2 | (I) | 3F |
(B) | 1 mol of MnO-4 to Mn2+ | (II) | 2F |
(C) | 1.5 mol of Ca from molten CaCl2 | (III) | 1F |
(D) | 1 mol of FeO to Fe2O3 | (IV) | 5F |
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
Organic Chemistry is a subset of chemistry dealing with compounds of carbon. Therefore, we can say that Organic chemistry is the chemistry of carbon compounds and is 200-225 years old. Carbon forms bond with itself to form long chains of hydrocarbons, e.g.CH4, methane and CH3-CH3 ethane. Carbon has the ability to form carbon-carbon bonds quite elaborately. Polymers like polyethylene is a linear chain where hundreds of CH2 are linked together.
Read Also: Organic Compounds
Organic chemistry is applicable in a variety of areas including-