Some Basic Concepts of Chemistry Test - 14

Equivalent weight of \(K_2Cr_2O_7\) is 49 g/mol. Here, 6 e- participate in the reaction, hence chemical equivalent is 6.

Equivalent weight of \(K_2Cr_2O_7\) in acidic medium is 49 g/mol.

(i) A chemical element is a substance that contains only one type of atom. If a substance contains more than one type of atom, it is a compound.

(ii) An element can be a solid, liquid or gas. The smallest particle of such an element is an atom.

Urea is a common source of nitrogen in all solid nitrogenous fertilizers and it is widely used as a nitrogen release fertilizer.

The standard crop-nutrient rating (NPK rating) of urea is 46-0-0.

Hence, it contains 46% elemental nitrogen (N) 0% elemental phosphorous (P), and 0% elemental potassium (K).

22400 ml is the volume of \(O_2\) at NTP = 32gm of \(O_2\)

1 ml is the volume of \(O_2\) at NTP = \(\frac{32}{22400}\)

112 ml is the volume of \(O_2\) at NTP = \(\frac{32}{22400}\times112\)

= 0.16g of \(O_2\)

All the zeroes between two non-zero digits are significant. Hence in 60.0001 significant figures is 6.

Equivalent weight = \(\frac{Molecular \;weight} {Valency}\)

Molecular weight of

\(C_2H_2O_4\) .\(2H_2O\) = \(\frac{126}{2}\) = 63

Sp. heat x atomic wt. = 6.4

0.16 x atomic wt.= \(\frac{6.4}{Atomic\; wt.}\)

= \(\frac{6.4}{0.16}\)

= 40.

From molarity equation

\(M_1V_1 + M_2 V_2\)

= \(MV_{(total)} 2 \times \frac{10}{1000} + 0.5 \times \frac{200}{1000}\)

= \(M \times \frac{210}{1000}\)

120 = M x 210 M

= \(\frac{120}{210}\)

= 0.57 M

(i) A student performs a titration with different burettes and finds titter values of 25.2 mL, 25.25 ml the number of significant figures in the average titter value is 3. This is according to the rules of significant figures.

(ii) If two or more numbers having different precision are to be added or subtracted, the answer should be taken as the number of decimal places in the number having less decimal places.

(i) As per the law of multiple proportions, when two elements combine to form two or more compounds, the ratios of the masses of two interacting elements in the two compounds are small whole numbers.

(ii) Thus, different proportions of oxygen in the various oxides of nitrogen prove the law of multiple proportions.

The definition of Avogadro's number of \(6.022 \times 10^{23}\)  mole is the number of atoms or molecules per one gram atomic weight.

1. Fractional distillation is employed for a separating mixture of two or more volatile liquids having boiling points close to each other.

2. Acetone has a boiling point 60°C, whereas methanol has a boiling point of 65°C. Hence, it can be separated by fractional distillation.

Trailing zeros in a whole number is not significant. Hence 3400 has 2 significant digits only.

The sample contains impurity. The impurity won't contribute to the normality of the solution.

So, we need to take more amount of samples than the theoretical weight so that the theoretical weight of compound equals weight of compound in the sample.

According to definition of molar solution → A molar solution is one that contains one mole of a solute in one litre of the solution.

\(2KMnO_4 + 3H_2SO_4\) → \(K_2SO_4 + 2MnSO_4 + 3H_2O + [O]\)

\(2FeSO_4 + H_2SO_4 + [O]\) → \(Fe_2(SO_4)_3 + H_2O] \times 5\)

[Mohr-salt]

\(2KMnO_4 + 10FeSO_4 + 8H_2SO_4\) → \(K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 8H_2O\)

Pressure = \(\frac{Force}{Area}\)

\(\frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]\)

= \(\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]\)

Energy per unit volume

\(\frac{[ML^2T^{-2}]}{[L^3]}\)

= \([ML^{-1}T^{-2}]\)

(i) All non-zero digits are significant.

(ii) Non-zero digits to the right of the decimal point are significant.

(iii) Zeroes to the left of the first non-zero digit in a number are not significant.

So, the number of significant figures for the numbers 161 cm, 0.161 cm and 0.0161 cm are same, i.e. 3.

(i) In case of Ultrafiltration, it is particle size of micron sizes present in liquid or gas is getting filtered.

(ii) In molecular sieving, permeability of molecular sieve membranes based on the size and shape of the molecules is employed.

(iii) In gravity separation difference in density of particles is employed

(iv) In molecular attraction, mainly exchange of cations to anions among given two molecules.

(v) In atomic absorption, elements are identified based on the absorption of optical radiation by free atoms in gaseous state.

Carbon is a chemical element with the symbol C and atomic number 6. It is nonmetallic and tetravalent-making four electrons available to form covalent chemical bonds. It belongs to group 14 of the periodic table.