Chemistry

The name of covalent and ionic bonds tell their formation story in small detail. Co- means sharing or coexisting, so whenever there is bond fromation due to sharing of valence electron,  it is known as covalent bond. Similerly, ionic bonds suggest that when a bond formation takes place due to ions, which attract each other, it is considered as ionic bond. Ionic bonds are also known as electrovalent bonds. There are some major differences between Electrovalent and covalent bonds. Chekc the table below:

For more detail related to the chemical bonding in class 11 chemistry read our notes.

Chemical Bonding and Molecular Structure

Ala – Gly – Leu-Ala- Val

Two amino acid units are connected by peptide linkage. So number of peptide linkage = 4.

Number of moles of gas = $\frac{22.4}{22400}=\frac{1}{1000}moles$

Mass of nitrogen = $\left(\frac{1}{1000}*28\right)gm$

% of N in organic compound = $\frac{\left(\frac{28}{1000}\right)}{0.2}*100\mathrm{\%}=14\mathrm{\%}$

In acidic solution

$3\stackrel{+6}{Mn}{O}_4^{2-}+4H^{+}\to 2\stackrel{+7}{Mn}{O}_4^{-}+\stackrel{+4}{Mn}{O}_2+2H_{2}O$

Difference in oxidation state of Mn is product

= 7 – 4 = 3

Baryte – BaSO₄ [Sulphate based ore]

Galena – PbS

Zinc blende – ZnS

Copper pyrite – CuFeS₂

Number of moles of CH₃COOH Adsorbed on 0.6 gm wood charcoal.

$=\left(0.2*200\right)*10^{-3}-\left(0.1*200\right)*10^{-3}$                

= 20 * 10^-3 moles

Mass of CH₃COOH absorbed = 20 * 10^-3 * 60 = 1.2 gm

Mass of CH₃COOH adsorbed per gram = $\frac{1.2}{0.6}=2g$ of charcoal.

$K=\frac{1}{R}*cellconstant$

= 0.152 * 10^-3 * 1750

 = 266 * 10^-3 cm^-1

$\begin{array}{l}PC{l}_5\left(g\right)?PC{l}_3\left(g\right)+C{l}_2\left(g\right)\\ t=05moles00\\ \underset{\_}{-xmole}\underset{\_}{+xmole}+\underset{\_}{xmole}\end{array}$

Eq.(5 – x) mole x molex mole

Total number of moles at equilibrium

$=\left(5-x\right)+x+x+n_{N_2}=\left(7+x\right)$

$K_p=\frac{\left(P_{PC{l}_3}\right)*{\left(PC{l}_2\right)}_{Cl}}{{\left(P_{PC{l}_5}\right)}_{eq}}=\frac{\left(\frac{3}{10}*2.46\right)*\left(\frac{3}{10}*2.46\right)}{\left(\frac{2}{10}*2.46\right)}=1.107atm$

= 1107 * 103 atm

According to Henry's law

$P_{Gas}=K_H*x_{Gas}$

$=0.835=1.67*10^3*\left[\frac{n_{C{O}_2}}{n_{C{O}_2}+55.5}\right]$

As $n_{C{O}_2}<<55.5$

$\Rightarrow n_{C{O}_2}=27.78*10^{-3}moles$

Mass of CO₂ gas = $n_{C{O}_2}*\left(MM\right)C{O}_2$

= 1222 * 10^-3 gm