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The atomic number of phosphorus is 15 and the 3p sublevel has three unpaired electrons.

solute + solvent = solution at 100oC; 75g + 100g = 175g at30oC; \(\frac{25g}{50g}\) + 100g = 125g From 100oC - 30oC; 175g CuSO4 ppt 50g From 100oC - 30oC; 50g CuSO4 ppt \(\frac{50 \times 50}{175}\) = 14.3

Q = mct = (50 x 4.18 x 3.4) / (1000) = 0.71606 kj 1.1 g of \(CaCl_2\) contains (1.1) / (111) = 0.01 moles of \(CaCl_2\) If 0.01 liberates 0.7106 quantity of heat. 1 will liberate (0.7106 x 1) / (0.01) Exothermic = 71.06 = -71.1

3 - \(\frac{1}{3}\) - (\(\frac{5}{4}\) x \(\frac{2}{3}\)) + 1\(\frac{2}{5}\) = \(\frac{10}{3}\) - \(\frac{5}{6}\) + \(\frac{7}{5}\) = \(\frac{100 - 25 + 42}{30}\) = \(\frac{117}{30}\) = 3.9 \(\approx\) 4

\(\begin{array}{c|c} 7056 \\ \text{- 4577} \\\hline 2257 \end{array}\) By trial and error method Let the base to 8 i.e. Let n = 8 and it is easily verified that the subtraction holds. The subtraction does not hold when other values of n are tried n = 8

(\(\frac{1}{0.05} \div \frac{1}{5.005}\)) - (0.05 x 2.05) (\(\frac{1}{0.05}\) x 5.005) - (0.05 x 2.05) \(\frac{5.005}{0.05}\) - 0.1025 100.1 - 0.1025 = 99.9975 99.998 to 3 decimal place.

\(\frac{62}{3}\) = 20.6666..... = 20.7 to 3 sig. fig.

P produces 20,000 bags P reduces production by 5% i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\) = 1000 Q produces 15000 bags Q increases production by 5% i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\) = 750 Effective loss by the two = 1000 - 750 = 250 bags

% of water in the mixture = \(\frac{\text{Total Amount of water}}{\text{Total quantity of spirit}}\) x \(\frac{100}{1}\) \(\frac{3(\frac{20}{100}) + 5 (\frac{15}{100})}{3 + 5}\) x \(\frac{100}{1}\) = \(\frac{\frac{6}{10} + \frac{75}{100}}{8}\) x \(\frac{100}{1}\) = \(\frac{0.6 + 0.75}{8}\) x \(\frac{100}{1}\) = \(\frac{1.35}{8}\) x \(\frac{100}{1}\) = \(\frac{33.75}{2}\) = 16.875 = 16\(\frac{7}{8}\)

2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9 [\(\frac{2}{5}\))2 x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\) = log \(\frac{72}{125}\) = log \(\frac{5}{2}\) = log \(\frac{10}{4}\) = log 10 - log 4 = log10 10 - log10 22 = 1 - 2 log2

\(\frac{1}{1 + \sqrt{5}}\) - \(\frac{1}{1 - \sqrt{5}}\) = \(\frac{1 - \sqrt{5} - 1 - \sqrt{5}}{(1 + \sqrt{5}) (1 - \sqrt{5}}\) = \(\frac{-2\sqrt{5}}{1 - 5}\) = \(\frac{-2\sqrt{5}}{- 4}\) = \(\frac{1}{2}\sqrt{5}\)

\(\frac{2\sqrt{3} + 3 \sqrt{2}}{3\sqrt{2} - 2 \sqrt{3}}\) = \(\frac{2\sqrt{3} + 3 \sqrt{2}}{3\sqrt{2} - 2 \sqrt{3}}\) x \(\frac{3\sqrt{2} + 2 \sqrt{3}}{3\sqrt{2} - 2 \sqrt{3}}\) \(\frac{4(3) + 9(2) + 2(6) \sqrt{6}}{9(2) - 4(3)}\) \(\frac{12 + 18 + 12\sqrt{6}}{1`8 - 12}\) = \(\frac{30 + 12\sqrt{6}}{6}\) = 5 + 2\(\sqrt{6}\)

(x2 - 3x + 1)(x - a) = x3 - 3x2 + x - ax2 + 3ax - a = x3 - (3 + a) x2 + (1 + 3a)x - a

\(\frac{xy^2 - x^2y}{x^2 - xy^1}\) = \(\frac{(-2)(3)^2 - (-2)^2(3)}{(-2)^2 - (-2)(3)}\) = \(\frac{-30}{10}\) = -3

g(x) = x2 + 3x + 4 = g(x + 1) = (x + 1)^2 + 3(x + 1) + 4 = x2 + 1 + 2x + 3x + 3 + 4 = x2 + 5x + 8 g(x + 1) - g(x) = x2 + 5x + 8 - (x2 + 3x + 4) = x2 + 5x + 8 - x2 + 3x + 4 = 2x + 4 = 2(x + 2)