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The furring of kettles or boilers - this is as a result of the decomposition of \(Ca(HCO_3)_2\) into \(CaCO_3\), and this\(CaCO_3\) coats the inside of a kettle or boiler used in heating water containing dissolved \(Ca(HCO_3)_2\). When a kettle has been used to boil temporary hard water for quite some time, the inner surfaces become coated with a white fur-like layer. This layer (furring of the kettle) is due to the gradual deposition of calcium trioxocarbonate(IV), from the decomposition of Calcium hydrogentrioxocarbonate (IV).

Zn\(^{2+}\) + 2e\(^-\) → Zn 2F will be liberated 65g of Zn, if 0.65gm of Zn is liberated, the quantity of electricity used X = (0.65/65) X 2 = 0.02F ∴ Hg\(^{2+}\) + 2e\(^-\) → Hg i.e 2F → 201gm of Hg ∴ 0.02 will liberated Y = \(\frac{0.02}{2}\) X 201 = 2.01g

When concentrated NaCl solution undergoes electrolysis, sodium hydroxide and hydrogen are produced at the cathode while chlorine is produced at the anode.

Increase in temperature will cause the forward reaction to occur, increasing the amounts of the products and decreasing the amounts of reactants.

The dehydration of tertiary alcohols is relatively simple and can be achieved by heating the alcohol to a high temperature or using a catalyst. The dehydrated product is a mixture of alkenes, with or without carbocation rearrangement. Recall that dehydration process involves removal of a molecule of water. As a result,the OH in the tertiary alcohol will combine with one atom of H from the CH\(_3\) attached to the C atom carrying the OH group, to form water, thereby introducing a C=C. The product of dehydrating a tertiary alcohol is an alkene. Therefore, option D is the correct answer.

In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS) In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s) ∴ ∠RQS = X + X = 2X Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS, 72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s) ∴ 72 = x + 2x 72 = 3x x = 72/3 x = 24o

r = \(\frac{2}{3}\)R ∴R = \(\frac{3}{3}\)R Area of small circle = πr2 = π(\(\frac{2R}{3}\))2 Area of the big circle πr2 = π\(\frac{(3R)^2}{3}\) Area of shaded portion = π(\(\frac{3R}{3}\))2 - π(\(\frac{2R}{3}\))2 = π[(\(\frac{3R}{3}\))2 - (\(\frac{2R}{3}\))2] = π[(\(\frac{3R}{3}) + (\frac{2R}{3}) - (\frac{3R}{3}\)) - (\(\frac{2R}{3}\))] = π[(\(\frac{5R}{3}\)) (\(\frac{R}{3}\))] = π x \(\frac{5R}{3}\) x \(\frac{R}{3}\) = 5/9πR2

Volume of cylinder = πr2h = π x 32 x 20 = π x 9 x 20 = 180 πcm3 Volume of hemisphere = 2/3πr3 = 2/3 x π x 32 = 2 x π x 32 = 2 x π x 9 = 18π = Volume of the solid = 180π x 18π = 198πcm3

∠PQR = 180 - 128 (∠s on a straight line) = 52o ∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s) ∠QPR + 128 = 180 ∠QPR = 180 - 128 = 52o ∴ΔPQR is an isosceles triangle

20 + X + 50 + 40 + 2X + 60 = 260 3X + 170 = 260 3X = 260 - 170 3x = 90 x = 30 Absent for at least 4 days 4 5 6 40 2x 60 i.e 40 + 2x + 60 = 40 + (2 x 30) + 60 = 40 + 60 + 60 = 160

Given, Length of minor arc = 5.5m Angle subtended by minor arc = 360^\circ - 210^\circ = 150^\circ \(\therefore 5.5 = \frac{150}{360} \times 2 \times \frac{22}{7} \times r \) \(\frac{55r}{21} = 5.5\) \(r = \frac{5.5 \times 21}{55}\) r = 2.1m Length of major arc = \(\frac{210}{360} \times 2 \times \frac{22}{7} \times 2.1\) = \(7.7m \approxeq 8m\) (to the nearest metre)

\(BD^2 = 4^2 + 4^2\) \(BD = \sqrt{16 + 16} = \sqrt{32}\) \(BD = 4\sqrt{2} cm\) \((2\sqrt{3})^2 = (2\sqrt{2})^2 + h^2\) \(h^2 = 12 - 8 = 4\) \(h = \sqrt{4} = 2 cm\) Volume of pyramid = \(\frac{a^2 h}{3}\) = \(\frac{4^2 \times 2}{3}\) = \(\frac{32}{3} = 10\frac{2}{3} cm^3\)

Curved surface area or a cone \(=\pi rl\) from the information \(l^2 = 4^2 + 3^2 = 16+9\\ l = \sqrt{25} = 5; ∴ C.S.A\hspace{1mm} = \frac{22}{7}\times 3 \times 5\\ Since \frac{22}{7}=\pi ∴ C.S.A\hspace{1mm} =\hspace{1mm}15\pi\)

< UQR = 29^\circ (alternate angles) < PQR = < PQU + < UQR = 36^\circ + 29^\circ = 65^\circ

\(5\frac{1}{4}\div \left(1\frac{2}{3}- \frac{1}{2}\right)\\ \frac{21}{4}\div \left(1\frac{4-3}{6}\right)\\ \frac{21}{4}\div \left(1\frac{1}{6}\right)\\ \frac{21}{4} \times \frac{6}{7}= 4\frac{1}{2}\)