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The order of reactivity of metals is as follows, Potassium > Sodium > Lithium > Calcium > Magnesium > Aluminium > Zinc > Iron > Copper > Silver > Gold.

In cathodic protection,iron object is made cathode by connecting it with reactive elements like magnesium,zinc.

For a diatomic divalent element like oxygen, the following reaction takes place: 2H\(_2\)0 → O\(_2\) + 4H\(^+\) + 4e\(^-\) ⇒ 4e\(^-\) are transferred to liberate 1 mole of O\(_2\) So, to calculate the volume of O\(_2\) 4 Faradays (4 x 96,500) of electricity to produce 22.4 dm\(^3\) of O\(_2\) at S.T.P 4 X 96,500 C = 22.4 dm\(^3\) 9650 C = x dm\(^3\) x = \(\frac{9650 X 22.4}{ 4 X 96,500}\) x = 0.56dm\(^3\)

R is concentrated hydrochloric acid and potassium tetraoxomanganate(VII) (See diagram above)

The concentration of the \(CaCO_3\) decreases as reaction time decreases. therefore option D is correct.

\(25^{x} + 3(5^{x}) = 4\) Let \(5^{x}\) = y. \((5^{2})^{x} + 3(5^{x}) - 4 = 0\) \(y^{2} + 3y - 4 = 0\) \(y^{2} - y + 4y - 4 = 0\) \(y(y - 1) + 4(y - 1) = 0\) \((y + 4)(y - 1) = 0\) \(y = -4 ; y = 1\) y = -4 is not possible. y = 1 \(\implies\) x = 0.

Let the sum of the 12 numbers be x and the 13th number be y. \(\frac{x}{12} = 3 \implies x = 36\) \(\frac{36 + y}{13} = 5 \implies 36 + y = 65\) \(y = 65 - 36 = 29\)

\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\) \(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} - \frac{10}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-1}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-2 + 15}{20})]\) = \(\frac{1}{3} \div [\frac{5}{7} \times \frac{13}{20}]\) \(\frac{1}{3} + [\frac{13}{28}]\) = \(\frac{1}{3} \times \frac{28}{13}\) = \(\frac{28}{39}\)

\(\frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4}\) = \(\frac{36}{420} \times \frac{54}{90} \times \frac{63}{240}\) = \(\frac{6}{70} \times \frac{18}{30} \times \frac{21}{80}\) = \(\frac{27}{2000}\) = 0.0135 \(\approx\) = 0.014

\(\frac{log_5 0.04}{log_3 18 - log_3 2}\) = \(\frac{log_5 0.04}{log_3(\frac{18}{2})}\) = \(\frac{log_5 0.04}{log_3 9}\) = \(\frac{-2}{2}\) = -1 Let log5 0.04 = x 5x = 0.04 x = \(\frac{4}{100}\) = 5\(^{-2}\) Let log\(_3\) 9 = z 3\(^2\) = 3\(^z\) z = 3

8x-2 = \(\frac{2}{25}\) = 200x-2 = 2 = 100x-2 = 1 x-2 = \(\frac{1}{100}\) x-2 = 10-2 x = 10

\(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\) Rearrange = \(\sqrt{48}\) + \(\sqrt{75}\) - \(\frac{9}{\sqrt{3}}\) = (\(\sqrt{16}\) x \(\sqrt{3}\)) + (\(\sqrt{25}\) x \(\sqrt{3}\)) - \(\frac{9}{\sqrt{3}}\) = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) - \(\frac{9}{\sqrt{3}}\) Rationalize \(\frac{9}{\sqrt{3}}\) \(\to\) = \(\frac{9}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = 3\(\sqrt{3}\) = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) - 3\(\sqrt{3}\) = 6\(\sqrt{3}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{\sqrt{2}}{2}\) = \(\frac{1.414}{2}\) = 0.707

A \(\subset\) B means A is contained in B i.e. A is a subset of B(A \(\cap\) B)' = A' A(A \(\cap\) B)' = A \(\cap\) A' The intersection of complement of a set P and P' has no element i.e. n(A \(\cap\) A') = \(\phi\)

\(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\) = \(\frac{2m - u + m - 2u)(2m - u - m + 2u)}{5(m + u)(m - u)}\) = \(\frac{3(m - u)(m + u)}{5(m + u)(m - u)}\) = \(\frac{3}{5}\)