CHEM 120 FINAL EXAM
6. (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)
7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant t...
CHEM 120 FINAL EXAM
6. (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. 01644446457Short16
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).
01644446459Short19
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) molarity = moles solute / liters solution
0.25 M = moles NaOH / 0.035 L
moles NaOH = 0.00875 moles NaOH 01644446450Short1
2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5) 3.5 pints is equivalent to 1656.116
1 pint = 473.176 ml
3.5 pints* 473.176mL = 1656.116mL 01644446452Short6 3. (TCO 3) What is the name of the following compound: Zn 3P2? (Points : 5) It's Zinc Phosphide 01644446453Short7
4. (TCO 3) What is the name of the following compound: AgNO 3? (Points : 5) Silver nitrate 01644446454Short11
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) Given that n = 2.78 mol; V = 4.25 L; and temperature [2.78 mol* 0.0821 L -atm/ mol-K*304 K)/4.25 L = 16.3 01644446456Short14 1. (TCO 7) (a, 5 pts) Given that the molar mass of H 3PO4 is 97.994 grams, determine the number of grams of H3PO4 needed to prepare 0.75L of a 0.25M H 3PO4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.25 M H 3PO4 solution can be prepared by diluting 50 mL of a 2.5M H3PO4 solution? Show your work. (Points : 10) Using the molar mass given, convert this amount to grams.
mass = 0.1875 mol * (97.994 g/mol) = 18.37 grams H3PO4
b. M1*V1 = M2*V2
w here: M1 = 0.25; V1 = ??; M2 = 2.5; V2 = 50 mL = 0.050 L
Solvig for V1:
0.25 * V1 = 2.5 * 0.050 V1 = 0.50 L
2. First convert the given mass of NaOH to volume (in mL) using the density of NaOH
Volume = 43 g * (1 mL/2.13 g) = 20.19 mL
Volume % = (volume of solute / volume of solution) * 100%
Volume % = (20.19 mL/120 mL) * 100% = 16.8 %
b. Volume % = volume of NaOH/ total volume
0.10 = 20.19 mL/total volume
Solving for total volume yields: 01644446463Essay3
2. (TCO 7) (a, 5 pts) What is the volume percent of a solution prepared by dissolving 21 g of NaOH in enough water to make a final volume of 120 mL? Show your work. (b, 5 pts) How many mL of a 10% solution can be made from the solution in part a? Show your work. (Points : 10) Part 1: based on the density of NAOH =2.1g/mL
then calculate the volume of 21g NaOHx(1mL/2.1g)=10 mL
so the percent volume is defined as % volume = (Volume of Solute/volume of solution) x 100
so w e have % volume = (10mL/120mL)x100=8.3%
Part 2: you cannot prepare 10 % (volume) from the above solution (8.3%) because the final solution is more concentrated than the initial 01644446464Essay5 01644446466Essay7
4. (TCO 11) Tungsten (W), with a mass number of 180 and an atomic number of 74, decays by emission of an alpha particle. Identify the product of the nuclear reaction by providing its atomic symbol (5 pts), mass number (5 pts), and atomic number (5 pts). (Points : 15) mass of 180 becomes 176
atomic number of 74 becomes 72
name: Hafnium
Symbol: Hf 01644446467Essay11
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