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Sixteen. B The oxidation-reduction response in a galvanic telephone happens spontaneously and gives a voltage that reasons a circulate of electrons, so ∆G° is unfavourable and E° is optimistic. additionally, from the equation ∆G° = –nFE°, we all know that ∆G° and E° regularly have contrary symptoms. we all know that Keq has to be more than 1 from both of the subsequent equations: ∆G° = –2. 303RT log ok, which tells us that if ∆G° is to be adverse, then log ok has to be optimistic, because of this okay has to be more than 1. Or: logK = , which tells us that if E° is to be optimistic, then logK has to be confident, which means ok has to be more than 1.
A) (B) (C) (D) (E) 19. 2 NO(g) + Br2(g) → 2 NOBr(g) For the response above, the experimental fee legislation is given as follows: cost = k[NO]2[Br2] Which of the statements under is correct concerning this response? (A) The response is first-order total. (B) The response is first-order with recognize to Br2. (C) The response is first-order with appreciate to NO. (D) The response is second-order total. (E) The response is second-order with recognize to Br2. 20. Which of the subsequent is the main most likely electron configuration of a sulfur atom in its flooring nation?
S is larger than 0. (A) II basically (B) III in simple terms (C) I and II in simple terms (D) I and III in basic terms (E) II and III simply 20. 2 S(s) + three O2(g) → 2 SO3(g) ΔH = +800 kJ/mol 2 SO3(g) → 2 SO2(g) + O2(g) ΔH = –200 kJ/mol in keeping with the knowledge given above, what's ΔH for the subsequent response? S(s) + O2(g) → SO2(g) (A) 300 kJ (B) 500 kJ (C) 600 kJ (D) 1,000 kJ (E) 1,200 kJ difficulties 1. power is published while glucose is oxidized within the following response, that is a metabolism response that happens within the physique.
2-molar CH3COOH and nil. 2-molar NaOH. half B Directions: Each of the questions or incomplete statements under is through 5 prompt solutions or completions. opt for the person who is better in each one case and fill within the corresponding oval at the resolution sheet. 15. A natural pattern of KClO3 is located to comprise seventy one grams of chlorine atoms. what's the mass of the pattern? (A) 122 grams (B) 170 grams (C) 209 grams (D) 245 grams (E) 293 grams sixteen. Which of the subsequent experimental approaches is used to split components via benefiting from their differing boiling issues?
Molecules with larger polarity may have larger dipole−dipole charm, so molecules with higher dipole moments are likely to have better melting and boiling issues. Dipole−dipole sights are quite vulnerable, even if, and those elements soften and boil at very low temperatures. such a lot components held jointly via dipole−dipole charm are gases or drinks at room temperature. London Dispersion Forces London dispersion forces ensue among impartial, nonpolar molecules. those very vulnerable points of interest take place as a result of random motions of electrons on atoms inside molecules.