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V = \sqrt2GM/R. =\sqrt[2 x 6.67 x 10\(^{-11}\) x 6.0 x 10\(^24\)] / 6.4 x 10\(^6\) = \sqrt12.506 x 10\(^7\) = 11,183.13 OR V =\sqrt2gR V=\sqrt[2 * 10 * 6.4 x 10\(^6\)] V = \sqrt128,000,000 11,313.708

16 days = 4 half-lives. The fraction left 16 days = after 4 half live, = (1/2)4 = (1)/16 Fraction that decayed = 1- \(\frac{1}{16}\) \(\frac{15}{16}\)

The correct answer is B. 2.0M of HCl. Here's why: - HCl (hydrochloric acid) is a strong acid, which means it completely dissociates in water to produce H\(^+\) ions and Cl\(^-\) ions. - The concentration of HCl is directly related to its ability to conduct electricity. A higher concentration of HCl means more ions are available to carry electrical charge. - Between options A and B, the 2.0M HCl solution has a higher concentration of ions, making it a better conductor of electricity. Ethanoic acid (CH\(_3\)COOH), on the other hand, is a weak acid, which means it only partially dissociates in water. As a result, it produces fewer ions and is a poorer conductor of electricity compared to HCl. So, even though option D has a higher concentration of ethanoic acid, it still won't conduct as much electricity as the 2.0M HCl solution.

During the electrolysis of an aqueous copper sulfate (CuSO\(_4\)) solution using platinum electrodes, copper ions (Cu\(^{2+}\)) are reduced to solid copper (Cu) at the cathode, and water (H\(_2\)O) is oxidized to oxygen gas (O\(_2\)) and hydrogen ions (H\(^+\)) at the anode. Here's a more detailed explanation: Cathode (Reduction): Copper ions (Cu\(^{2+}\)) from the solution migrate towards the cathode and gain two electrons (2e-) to form solid copper, which deposits on the electrode. Reaction: Cu\(^{2+}\)(aq) + 2e- → Cu(s) Anode (Oxidation): Water molecules (H\(_2\)O) are oxidized at the anode, losing electrons and forming oxygen gas (O\(_2\)) and hydrogen ions (H+). Reaction: 4OH\(^-\) + 4e\(^-\)(l) → 2H\(_2\)O(g) + O\(_2\)

For a reaction to be spontaneous, ∆G = - ve; ∆H = - ve; ∆S = + ve Where ∆G = ∆H - T∆S Given data; ∆G = ? ∆H = - 72KJ/mol, T = (25+273)K = 298K ∆S = -106J/mol or - 0.106KJ/mol/K : ∆G = -72 -(298 * - 0.106) = - 72 + 31.588 ∆G = - 40.412KJ/mol One of the condition for spontaneity of a reaction is ∆G to be negative.

Carbon exists in the combined state as petroleum, wood and natural gases.

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0. First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1. Then, a\(_0\) which is the coefficient void of x = 6. The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6. The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\). = \(\pm (1, 2, 3, 6)\). Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0. Therefore x = 1 is a root of the equation. Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6. x\(^2\) - x - 6 = (x - 3)(x + 2). x = -2, 3. \(\therefore\) The roots of the equation = 1, -2 and 3.

English = 52 students Health education = 36 students Total = 88 students = \(\frac{88}{160} \times 100%\) = 55%

IV is correct.

When the phrase 'a number of' is used in a sentence, it is followed by a plural verb, hence 'have been' is the correct answer.

The contracted word with a subject is 'He has'. Has is usually not contracted except it comes before 'been'

It is grammatically correct to say 'a dangerously low level'

The past perfect tense 'had been hoping' complete the sentences appropriately.

Would is the past tense of 'will', therefore 'applied' is the correct answer as it is in its past tense.

Ruthless means being cruel and heartless.