CHEM120 finals QUESTIONS AND CALCULATED ANSWERS GRADED A
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. (TCO 6) A gas at a temperature of 95 degrees C occupies a volume of 165 mL. Assuming constant pressure, determine the volume at 25 degrees C. Show your work. (Points : 5) Using Charles’ Law, (V1... /T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t1 +273) Thus, T1 = 95 + 273 = 368. We have V1 (165 mL) & T2 = (25 + 273) = 298. V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL. 7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant temperature, determine the volume of helium at 745 mmHg. Show your work. (Points : 5) 1021mL * 719 mm/745 mmHg = 985.36mL =985mL Using Boyle’s law, P1V1 = P2V2. We have V1 (1021 mL), P1 (719 mmHg) and P2 (745 mmHg). 8. (TCO 12) If one strand of a DNA double helix has the sequence T T A G C G A C G C, what is the sequence of the other DNA strand? (Points : 10) A A T C G C T G C G 1. (TCO 8) 35.0 mL of 0.25 M NaOH is neutralized by 23.6 mL of an HCl solution. The molarity of the HCl solution is (show your work): (Points : 5) 2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5) 3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5) 0 1644446457 Short 16 0 1644446459 Short 19 molarity = moles solute / liters solution 0.25 M = moles NaOH / 0.035 L moles NaOH = 0.00875 moles NaOH 0 1644446450 Short 1 3.5 pints is equivalent to 1656.116 1 pint = 473.176 ml 3.5 pints* 473.176mL = 1656.116mL 0 1644446452 Short 6 4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5) 5. (TCO 6) Calculate the pressure, in atmospheres, of 2.78 mol CO(g) in a 4.25 L tank at 51 degrees C. (Points : 5) 1. (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine the number of grams of H3PO4 needed to prepare 0.75L of a 0.25M H3PO4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.25 M H3PO4 solution can be prepared by diluting 50 mL of a 2.5M H3PO4 solution? Show your work. (Points : 10 [Show More]
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