Hard water is water that does not give lather easily with soap because it contains dissolved calcium and magnesium salts. Temporary hardness is due to bicarbonates of Ca and Mg (Ca(HCO3)2, Mg(HCO3)2) and can be removed by boiling or by Clark's method (adding calculated slaked lime). Permanent hardness is due to chlorides and sulphates of Ca and Mg and is removed by adding washing soda (Na2CO3) or by the ion-exchange (permutit/resin) method.

Solubility of hydroxides increases down the group (Mg(OH)2 is sparingly soluble while Ba(OH)2 is fairly soluble) because lattice energy decreases faster than hydration energy. Solubility of sulphates decreases down the group (MgSO4 is soluble but BaSO4 is almost insoluble) because hydration energy decreases faster for the larger ions. Flame colours: calcium gives brick-red, strontium gives crimson-red and barium gives apple-green.

Diborane (B2H6) has two boron atoms and six hydrogen atoms. Four hydrogen atoms (two on each boron) are terminal and are bonded by normal two-centre two-electron (2c-2e) B-H bonds. The remaining two hydrogen atoms are bridging: each forms a three-centre two-electron (3c-2e) B-H-B bond, often called a banana bond. Each boron is sp3 hybridised. The molecule is electron-deficient because it does not have enough valence electrons for normal two-electron bonds throughout.

B2H6 (2 bridging + 4 terminal H)

Structural (constitutional) isomerism occurs when compounds have the same molecular formula but different arrangements of atoms. Three types: (1) Chain isomerism - different carbon skeletons, e.g. n-butane and isobutane (C4H10). (2) Position isomerism - same skeleton but a substituent or functional group in different positions, e.g. 1-propanol and 2-propanol. (3) Functional isomerism - same molecular formula but different functional groups, e.g. ethanol (alcohol) and dimethyl ether (C2H6O).

Hyperconjugation is the delocalisation of the sigma electrons of a C-H bond (adjacent to a positively charged carbon or a double bond) into the empty p-orbital or pi-system; it is also called no-bond resonance. In carbocations, the more alkyl groups attached to the positive carbon, the more C-H bonds are available for hyperconjugation, so more delocalisation of charge occurs. Hence stability order is tertiary > secondary > primary > methyl carbocation.

Crystallisation: used to purify solids; the impure solid is dissolved in a suitable hot solvent, filtered, and cooled so that the pure compound crystallises out while soluble impurities remain in solution. Simple distillation: used to separate a volatile liquid from a non-volatile impurity or two liquids with a large difference in boiling points; the liquid is boiled and the vapour is condensed and collected. Steam distillation: used to purify liquids that are steam-volatile and immiscible with water; steam is passed through the mixture so the compound distils over below its normal boiling point (e.g. aniline).

Ethene is prepared by heating ethanol with concentrated sulphuric acid at about 443 K (170 deg C), which removes a molecule of water (dehydration): CH3CH2OH -> CH2=CH2 + H2O. Test for unsaturation: (1) alkenes decolourise reddish-brown bromine water (Br2/H2O). (2) alkenes decolourise cold dilute alkaline KMnO4 (Baeyer's reagent), turning the purple colour colourless and forming a diol. These tests confirm a carbon-carbon double bond.

CH3CH2OH -> CH2=CH2 + H2O

Huckel's rule states that a planar, cyclic, fully conjugated ring is aromatic if it contains (4n + 2) pi electrons, where n = 0, 1, 2, 3... Benzene is aromatic because it is planar and cyclic, every carbon is sp2 hybridised with continuous conjugation (a delocalised pi system), and it has 6 pi electrons, which fits (4n + 2) with n = 1. This delocalisation gives benzene extra stability (resonance/aromatic stabilisation).

(4n + 2) pi electrons

Ethane (alkane) does not react with bromine water or Baeyer's reagent (no decolourisation) as it is saturated. Ethene (alkene) decolourises both bromine water and cold dilute alkaline KMnO4 (Baeyer's reagent) but gives no precipitate with ammoniacal silver nitrate. Ethyne (terminal alkyne) also decolourises bromine water and Baeyer's reagent, and in addition forms a white precipitate of silver acetylide with ammoniacal silver nitrate (and a red precipitate with ammoniacal cuprous chloride), because of its acidic terminal hydrogen.

n(T) = 150, n(C) = 225, n(T intersection C) = 100. Students drinking at least one = n(T union C) = n(T) + n(C) - n(T intersection C) = 150 + 225 - 100 = 275. Students drinking neither = total - n(T union C) = 600 - 275 = 325.

neither = n(U) - n(A union B)

A union B = {2,3,4,5,6,7,8}, so (A union B)' = U - (A union B) = {1,9,10}. Now A' = {1,6,7,8,9,10} and B' = {1,2,3,9,10}, so A' intersection B' = {1,9,10}. Since (A union B)' = {1,9,10} = A' intersection B', De Morgan's law is verified.

(A union B)' = A' intersection B'

For f(x) to be real, the expression under the square root must be non-negative: 9 - x^2 >= 0, i.e. x^2 <= 9, so -3 <= x <= 3. Hence the domain is [-3, 3]. As x varies over this interval, 9 - x^2 ranges from 0 (at x = +/-3) to 9 (at x = 0), so sqrt(9 - x^2) ranges from 0 to 3. Hence the range is [0, 3].

sqrt() defined only for non-negative values

f(2) = 2^2 + 1 = 4 + 1 = 5. f(-1) = (-1)^2 + 1 = 1 + 1 = 2. For f(x) = 10: x^2 + 1 = 10, so x^2 = 9, giving x = +3 or x = -3.

f(x) = x^2 + 1

360 revolutions per minute = 360/60 = 6 revolutions per second. One revolution corresponds to an angle of 2pi radians. Therefore the angle turned in one second = 6 x 2pi = 12pi radians.

angle = number of revolutions x 2pi

Expand each factor: cos(A + B) = cos A cos B - sin A sin B and cos(A - B) = cos A cos B + sin A sin B. Their product is of the form (p - q)(p + q) = p^2 - q^2 with p = cos A cos B and q = sin A sin B. So it equals cos^2 A cos^2 B - sin^2 A sin^2 B. Writing cos^2 B = 1 - sin^2 B and sin^2 A = 1 - cos^2 A: = cos^2 A (1 - sin^2 B) - (1 - cos^2 A) sin^2 B = cos^2 A - cos^2 A sin^2 B - sin^2 B + cos^2 A sin^2 B = cos^2 A - sin^2 B. Hence proved.

cos(A+B)cos(A-B) = cos^2 A - sin^2 B

Using sin^2 x + cos^2 x = 1: sin^2 x = 1 - (9/25) = 16/25, so sin x = +/- 4/5. In the second quadrant sine is positive, so sin x = 4/5. Then tan x = sin x / cos x = (4/5) / (-3/5) = -4/3.

sin^2 x + cos^2 x = 1

The principal value for which sin x = 1/2 is x = pi/6. The general solution of sin x = sin a is x = n pi + (-1)^n a, where n is any integer. Here a = pi/6, so the general solution is x = n pi + (-1)^n (pi/6), n belongs to the set of integers.

sin x = sin a => x = n pi + (-1)^n a

The principal value for which tan x = sqrt(3) is x = pi/3. The general solution of tan x = tan a is x = n pi + a, where n is any integer. Here a = pi/3, so the general solution is x = n pi + pi/3, n belongs to the set of integers.

tan x = tan a => x = n pi + a

Write sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x. Substitute sin 2x = 2 sin x cos x and cos 2x = 1 - 2 sin^2 x: = (2 sin x cos x) cos x + (1 - 2 sin^2 x) sin x = 2 sin x cos^2 x + sin x - 2 sin^3 x. Replace cos^2 x = 1 - sin^2 x: = 2 sin x (1 - sin^2 x) + sin x - 2 sin^3 x = 2 sin x - 2 sin^3 x + sin x - 2 sin^3 x = 3 sin x - 4 sin^3 x. Hence proved.

sin 3x = 3 sin x - 4 sin^3 x

Modulus |z| = sqrt(1^2 + (sqrt3)^2) = sqrt(1 + 3) = sqrt4 = 2. For the argument, tan(theta) = sqrt3/1 = sqrt3, and since both real and imaginary parts are positive (first quadrant), theta = pi/3. Polar form: z = 2(cos(pi/3) + i sin(pi/3)).

z = r(cos theta + i sin theta)