Chemistry Test-4

A reversible galvanic cell is connected to an external battery. If the EMF of the battery is less than the EMF of the galvanic cell, then the current will flow from the galvanic cell to the battery.

The atoms of elements belonging to the same group of periodic table have the same number of valence electrons.

The elements in the periodic table are arranged according to the Modern periodic law. It states that the physical and chemical properties of the elements are periodic functions of their atomic numbers. The physical and chemical properties depend on the number of valence electrons in an atom. These valence electrons are responsible for the vast majority of physical properties, chemical properties, and reactivities of the elements.

Antibonding molecular orbitalshave electron density which is in between the atoms.

In chemical bonding theory, an antibonding orbital is a type of molecular orbital (MO) that weakens the chemical bond between two atoms and helps to raise the energy of the molecule relative to the separated atoms. Such an orbitals:

• have electron density which is not between the atoms (there is a node between the atoms).
• are higher in energy than the original atomic orbitals.
• if these orbitals are occupied by electrons, it pulls the atoms away from each other and weakens or prevents bonding.

Anti-bonding orbitals are formed where the atomic orbitals combine such that it leads to predominantly destructive interference i.e. subtraction of wave function. The most important feature of antibonding orbitals is that the molecular orbitals have higher energy than the corresponding atomic orbitals.

Carboxylationis not a process of formation of alkanes.

Carbonylation (carboxylation) is the combination of an organic compound with carbon monoxide and carbonylation refers to reactions that introduce carbon monoxide into organic and inorganic compounds.

Alkanes can be prepared from alkyl halides (except fluorides) through reduction with zinc and dilute hydrochloric acid.

\(\ce{CH _{3}- Cl + H _{2} \rightarrow CH _{4}+ HCl}\)

Alkane can be prepared from alkene and alkyne through the process of hydrogenation.

Wurtz reaction: In dry ethereal solution, on treating alkyl halides with sodium metal, roduction of alkanes is higher. By this reaction, we can achieve higher alkanes with an even number of carbon atoms.

\(\ce{CH _{3}- Br +2 Na + BrCH _{3} \rightarrow CH_{3}- CH_{3} +2NaBr}\)

The reaction, in which anions and cations of two different molecules exchange places, forming two completely different compounds, is called double displacement reaction. The chemical bonds between the reactants may be either covalent or ionic. A double displacement reaction is also called a double replacement reaction, salt metathesis reaction, or double decomposition. The reaction between silver nitrate and sodium chloride is a double displacement reaction. The silver gives its nitrite ion for the sodium's chloride ion, causing the sodium to pick up the nitrate anion.

\(\mathrm{AgNO}_{3}+\mathrm{NaCl} \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_{3}\)

Considering \(+\mathrm{I}\) effect of alkyl groups, \(\mathrm{NH}_{3}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), and \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) can be arranged in the increasing order of their basic strengths as:

\(\mathrm{NH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\)

Again, in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) acceptability of proton is less than \(\mathrm{NH}_{3}\). Thus, we have:

Aromatic amine \(<\) Aliphatic Amine (Basic character)

Also due to the \(-I\) effect of \(\mathrm{C}_{6} \mathrm{H}_{5}\) group, the electron density on the \(\mathrm{N}\)-atom in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}\) is lower than that on the \(\mathrm{N}\)-atom in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), but more than that in \(\mathrm{NH}_{3}\). Therefore, the given compounds can be arranged in the order of their basic strengths as:

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{NH}_{3}<\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) (overall basic strength)

Dextrose does not represent a disaccharide.

Disaccharides are the smallest and commonest oligosaccharides found in nature. They are formed by the condensation of two monosaccharide molecules. The most important disaccharides are sucrose, lactose, and maltose.

Hard water contains calcium and magnesium ions. These ions form insoluble calcium and magnesium soaps respectively when sodium or potassium soaps are dissolved in hard water.

\({ }^7 C _{17} H _{35} COONa + CaCl _2 \rightarrow 2 NaCl +\left( C _{17} H _{35} COO \right)_2 Ca\)

These insoluble soaps separate as scum in water and are useless as cleansing agent. In fact these are hinderance to good washing, because the precipitate adheres onto the fibre of the cloth as gummy mass.

Given:

Substitute 5 g for a mass of solute and 295 g for the mass of solvent and calculate the mass of solution.

As we know,

Mass of solution = mass of solute + mass of solvent

Mass of solution = 5 g +295 g = 300 g

Now, we have a mass solution and also we have given the density of the solution so, calculate the volume of the solution as follows:

Density \(=\frac{\text { mass }}{\text { volume }}\)

volume \(=\frac{\text { mass }}{\text { Density }}\)

Substitute 300 g for a mass of solution and 1.5g/cc for the density of the solution and calculate the volume of the solution.

volume of solution \(=\frac{300 \mathrm{~g}}{1.5 \mathrm{~g} / \mathrm{cc}}\)

volume of solution \(=200 \mathrm{cc}\)

Now, to calculate the moles of solute using the volume of solution and molarity of solution convert the volume of solution in L.

\(1 \mathrm{~L}=1000 \mathrm{cc}\)

\(200 \mathrm{cc} \times \frac{1 \mathrm{~L}}{1000 \mathrm{cc}}=0.2 \mathrm{~L}\)

Substitute 0.05 M for molarity and 0.2 L for the volume of solution and calculate the moles of solute as follows:

Molarity \(=\frac{\text { moles of solute }}{\mathrm{L} \text { of solution }}\)

Moles of solute \(=\) Molarity \(\times \mathrm{L}\) of solution

Moles of solute \(=0.05 \mathrm{M} \times 0.2 L=0.01 \mathrm{~mol}\)

Now, we have moles of solute and mass of solute so calculate the molecular weight of solute as follows:

molecular weight \(=\frac{\text { mass }}{\text { mole }}\)

Substitute 0.01 mol for moles of solute and 5 g for a mass of solution and calculate the molecular weight of the unknown solute.

Molecular weight \(=\frac{5 \mathrm{~g}}{0.01 \mathrm{~mol}}\)

Molecular weight \(=\frac{5 \mathrm{~g}}{0.01 \mathrm{~mol}}\)

Molecular weight \(=500 \mathrm{~g} / \mathrm{mol}\)

Thus, the molecular weight of the unknown solute is 500 g/mol.