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The overall order of a reaction is the sum of the exponents in the rate law. The rate equation is \(\frac{-\Deltax}{\Deltat}\) = K[X]\(^2\)[Y]. The exponent for [X] = 2 The exponent for [Y] = 1 The overall order is the sum of the exponents = 2 + 1 = 3

Like aqueous mineral acids, an alkanoic acid like propanoic acid (CH\(_3\)CH\(_2\)COOH) neutralizes a base like NaOH to produce a salt and water only in a NEUTRALIZATION REACTION.

Chlorine and hydrogen do not react vigorously without heating unless exposed to sunlight or a flame. The reaction is spontaneous and explosive when exposed to these conditions. However, chlorine, bromine, and iodine all dissolve in alkali to form alkali metal halides. Alkali metals are electropositive elements that bond ionically with halogens, which are electronegative elements.

The probabilities of the three boys winning are independent and are given as: - Boy A: \( \frac{1}{2} \), - Boy B: \( \frac{1}{3} \), - Boy C: \( \frac{1}{4} \) "One and only one wins" means exactly one boy wins (the other two lose). The three possible cases are: A wins, B loses, C loses \( P = \frac{1}{2} \times \left(1 - \frac{1}{3}\right) \times \left(1 - \frac{1}{4}\right) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} \) B wins, A loses, C loses \( P = \left(1 - \frac{1}{2}\right) \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} \) C wins, A loses, B loses \( P = \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{3}\right) \times \frac{1}{4} = \frac{1}{2} \times \frac{2}{3} \times \frac{1}{4} = \frac{2}{24} \) Add them up: \( \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24} \)

Q + 2P = Q + 2(Q + P - Q) = 2(Q + P) - Q Q + P = \(\begin{pmatrix} 7 & -2 \\ 6 & 8 \end{pmatrix}\) 2(Q + P) = \(\begin{pmatrix} 14 & -4 \\ 12 & 16 \end{pmatrix}\) Q + 2P = 2(Q + P) - Q = \(\begin{pmatrix} 14-6 & -4-0 \\ 12-4 & 16-5 \end{pmatrix} = \begin{pmatrix} 8 & -4 \\ 8 & 11 \end{pmatrix}\) Determinant: |Q + 2P| = 8 \(\times\) 11 - (-4)\(\times\) 8 = 88 + 32 = 120.

Numbers: 3, 6, x, 7, 5 Mean = \(\frac{(21 + x)}{5}\) Standard deviation = 2, so the sum of squared deviations = 20. Solving: 2x² - 21x + 27 = 0 x = 9 or x = 1.5 ≈ 2 The **least possible value of x is 1.5

\DeltaABE and \DeltaACD are similar therefore \(\frac{ x}{ x + 4} = \frac{3}{6}\) 6x = 3(x+4) 6x = 3x + 12 3x = 12 x = 4 In \DeltaACD; \(L^2\) = \((4+4)^2 + 6^2\) = \(8^2 + 6^2\) = 64 + 36 = 100 L = \(\sqrt{100}\) = 10cm In ABE, \((AE)^2 = 4^2 + 3^2\) = 16 + 9 = 25 AE = 5cm ∴L = 10 - 5 = 5cm

% of those above 15 but less than 21 years age = \(\frac{25}{50}\) x \(\frac{100}{1}\) = 50%

If the side = 10 cm, correct to the nearest whole number, then The side ranges from 9.5 cm to 10.5 cm. (9.5 \(\leq\) s < 10.5) Perimeter of a rhombus = 4 x side \(\therefore\) 4 x 9.5 \(\leq\) perimeter < 4 x 10.5 = 38 cm \(\leq\) P < 42 cm

log\(_7\) 8 - log\(_7\) 2 + log\(_7\) 4 = log\(_7\) (8/2 x 4) = log\(_7\) 16 = 4 log\(_7\) 2

\(K\sqrt{28}+\sqrt{63}-\sqrt{7}=0\\ 2K\sqrt{7}+3\sqrt{7}-\sqrt{7}=0\\ 2K\sqrt{7}=-2\sqrt{7}\\ K=\frac{-2\sqrt{7}}{2\sqrt{7}}\\ K=-1\)

\(A = {3, \sqrt{2}, 2\sqrt{3}, \sqrt{9}, \sqrt{7}}\) n(A) = 5 Let the rational nos = R n(R) = 2 (3, \(\sqrt{9}\)) P(R) = 2/5

Total angle in a circle = 360^\circ \(\therefore\) 105 + 75 + 2k + k + 3k = 360^\circ 6k = 360 - 180 = 180 k = 180/6 = 30^\circ

Let x represent the total number of hours spent per week ∴ \(\frac{75}{360} \times x = \frac{5}{2}\) ∴ x =\(\frac{360 \times 5}{725 \times 2}\) =12 hours

Let the number of people that speak both English and French = x Then (7 - x) + x + (6 - x) = 11 13 - x = 11 \(\implies\) x = 2. \(\therefore\) P(picking a person that speaks both languages) = 2/11