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The basic function of microscope is magnification

Shortsightedness is corrected using a concave (curved inwards) lens which is placed in front of a myopic eye

Electrical work W = QE, Where Q is the quantity of change and E is the electromotive force. ∴ E = W = FxDQ_It Where D is the displacement,I = current and t = time But dimension of force, = unit of mass x unit of Acceleration = unit of mass xunit of velocityunit of time = unit of mass x unit of displacement__unit of time x unit of time ML/T2 = MLT-2 ∴ Dimension of emf E = MLT-2L = ML2T-3I-1__IT

% of H2 = 100 - 92.3 = 7.7% mole = Carbon [92.3 / 12] = 7.7 Hydrogen [7.7 / 1] = 7.7 divide with the lowest: Carbon - 7.1 / 7.1 = 1 Hydrogen - 7.1 / 7.1 = 1 empirical formula = CH (CH)n = 26 (12 + 1)n = 26 (13)n = 26 n = 26/13 = 2 molecular formula = C\(_2\)H\(_2\)

element : C | H Atomic Mass : 12.011 | 1.008 mass in 100 g of compound 88.9 | 11.100 we calculate the number of moles by dividing the mass by the atomic mass # moles in 100 g : 7.40155 | 11.01190 molar ratio 7.40155 : 11.01190 smallest No : 7.4015 divide by the smallest No. : 1.000 : 1.488 empirical formula is CH\(_2\)

An insoluble salt can be prepared by a precipitation method through a double decomposition reaction. Two soluble metal salts react with each other to form one soluble salt and one insoluble salt . Lead nitrate solution and sodium chloride solution react to produce solid lead chloride , leaving soluble sodium nitrate in solution. The lead chloride can be recovered by filtration.

Potash alum [kAl(S\({O}_4\))\(_2\)] n sodium aluminate (ii) (NaHCO\(_2\)) are added to cause coagulation. As fine particles to clump together, forming large particles heavy enough to settle at the bottom of the tanks as impurities.

Hydrogen fluoride is formed through covalent bonding (electron sharing) between a hydrogen atom and a fluorine atom. The reason is the strong electronegativity of the fluorine atom, which causes hydrogen and fluorine atoms to form hydrogen bonds in the liquid phase.

It does not conduct electricity. Every atom in a diamond is bonded to its neighbours by four strong covalent bonds, leaving no free electrons and no ions .

Tan 20º = \(\frac{68m}{x}\) x tan 20º = 68 x = \(\frac{68}{tan 20}\) = \(\frac{68}{0.364}\) x = 186.8 = 187m

x% of 240 = 12 \(\frac{x}{100} \times 240 = 12\) x = \(\frac{12 \times 100}{240}\) x = 5

\(\frac{(3.2)^2 - (4.8)^2}{3.2 + 1.8} = \frac{(3.2 - 4.8)(3.2 + 4.8)}{(3.2 + 4.8)}\) = 3.2 - 4.8 = -1.60

\(\sqrt{50} + \frac{10}}{\sqrt{2}} = \(\frac{\sqrt{50}}{1} + \sqrt{10}{\sqrt{2}}\) = \(\frac{\sqrt{50 \times 2} + 10}{\sqrt{2}}\) = \(\frac{\sqrt{100} + 10}{\sqrt{2}}\) = \(\frac{10 + 10}{\sqrt{2} = \frac{20}{\sqrt{2}}\) = \(\frac{20}{\sqrt{2}}\) \times \frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{20\sqrt{2}}{2}\) = 10\(\sqrt{2}\)

I = \(\frac{PRT}{100}\) where r = r% p.a; I = 0.36p 0.36p = \(\frac{P \times r \times 4}{100}\) \(\frac{0.36 \times 100}{4}\) = r r = 9

p - 2g + 1 = g + 3p.........(1) p - 2 = 0 .........(2) From (2), p = 2; put p = 2 into (1); 2 - 2g + 1 = g + 3(2) 3 - 2g = g + 6 -2g - g = 6 - 3 -3g = 3 g = \(\frac{3}{-3}\) g = -1