5: Reactions in Aq. Solutions – Flashcards
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| Electrolytes |
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| compounds whose aqueous solutions conduct electricity |
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| Strong electrolytes |
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| Substances whose solutions are good electrical conductors (such as sodium chloride) |
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| weak electrolytes |
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| poor electrical conductors (such as acetic acid) |
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| solution |
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| homogeneous mixture of two or more substances |
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| solvent |
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| the component of a solution present in greater quantity |
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| solute |
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the component of a solution present in lesser quantity. the solute is said to be dissolved in the solvent |
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| electrolyte |
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| a substance whose aqueous solutions contain ions. often these are ionic compounds |
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| nonelectrolyte |
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| a substance whose aqaueous solution does not contain ions. often these are molecular compounds |
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| strong electrolyte |
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| solutes that exist in solution completely or nearly completely as ions |
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| weak electrolyte |
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| solutes that exist in solution mostly in the form of molecules with only a small fraction in the forms of ions |
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| chemical equilibrium |
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| equal rates of the forward reaction of reactans converting into products and the reverse reaction of products reconverting back to reactants |
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| How many mL of 1.50 M KOH solution are needed to provide 0.125 mol of KOH? |
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| 0.125mol KOH x 1 L solution x 1000 mL = 83.3 mL solution 1.50 mol KOH 1 L solution |
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| You have a stock solution of 14.8 M NH3. how many mL of this colution should you dilute to make 100.0 mL of 0.25 M NH3? |
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| 100mL x 1 L x 0.25 M NH3 x 1 L stock solution x 1000 mL = 1.69 mL 1000 mL 1 L 14.8 mol NH3 1 L ss. ^# moles needed for final sol.^ ^mL of ss. w/ desired # moles^ |
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| glacial acetic acid is a liquid with a density of 1.049 g/mL at 25 C. calculate the molarity of a solution of acetic acid made by desolving 20.0 mL of glacial acetic acid at 25 C in enough water to make 250.0 mL of solution. |
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| 20.0 mL glacial x 1000 mL x 1.049 g acetic acid x mol CH3COOH =1.4 M 250.0 mL 1 L 1 mL glacial 60.0520 g CH3COOH |
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| calculate the KE in J of a 45 g of golf ball moving at 61 ms |
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| E= 1 mv2 2 E = 1 (45g) x kg x(61m/s)2 =84 J 2 |
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convert 84 J to calories |
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| 84 J x 1 cal = 20 cal 4.184 J |
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| A system releases 113 kJ of heat to the surroundings and does 39 kJ of work on the surroundings, endothermic or exothermic? |
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| <>E = q + w = 1.62 kJ - 0.874 kJ = 0.75 kJ, endothermic (q>0) |
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2 Mg(s) + O2(g) ----> 2 MgO(s) <> h = -1204 kJ |
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(a) exothermic (<>H<0) |
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| the specific heat of copper metal is 0.385 J/g K. How many J of heat are necessary to raise the temperature of a 1.42 kg block of coppter from 25.0 C to 88.5 C? |
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| <>T = 88.5 C -25.0 C = 63.5 C => 63.5 k 1.42 kg Cu x 1000 g Cu x 0.385 J x 63.5 k = 34700 J 1 g g k |
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| a 2.20 g sample of guinine (C6H4O2) is burned in a bomb calorimeter where total hear capacity is 2.854 kJ/ C. the temperature f the calorimeter increases from 23.44 C to 30.57 C. what is the heat capacity per gram of guinine? |
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| <>T = 30.57 C -23.44 C = 7.13 C q = 7.854 kJ/ C x 7.13 C x 1 = 25.45 kJ/g guinine 2.20 g 25.45 kJ x 108.0948 g guinine = 2750 kJ/mol guinine g guinine mol guinine |
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| calculate the wavelength of the radiation released when an electron moves from n= 6 to n = 2 |
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| E = h(freq.) = hc => wavelength = hE wavelength E(<>E=Ef-Ei=E(n=2)-E(n=6)) (6.06x10-34Js)(2.9979x108 ms-1) = -4.10x10-7m (-5.45x10-19J) - (-6.06x10-20J) wavelength = -410 nm, (neg. sign means radiation emitted) wavelength(emitted) = 410nm |
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| Heisenberg's uncertaint principle: a 1.50 mg mosquito moving with a speed of 1.40 m/s and the speed is known to + 0.01 m/s |
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| (;;x)(;;mv);;;; h;;;;;;;;; -or-; (;;x) ;;;;;;;;;; h;;;;;;;; ;;;;;;;;;;;;;;;;;;;;;; 4(pie);;;;;;;;;;;;;;;;;;;;;;;;;; 4(pie)(;;mv) ;;x ;;;;;;;;;;; (6.626x10-34 Js);;;;;;;;;x 106 mg ;;;;;;;;;; 4(pie) (1.50 mg) (0.01 m/s);;;;;;; ;kg ;;x; 3.52x10-27 m |
