Week 4 Text Exercises-Part III

Week 4 Text Exercises-Part III - Week 4 Text Exercises Part...

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Week 4 Text Exercises, Part III Prepare written answers to the following assignments from Introductory Chemistry: Exercise 50 in Ch. 13 A saline solution contains 1.1% NaCl by mass. How much NaCl is present in 87.2 g of this solution? Answer: 1.1 g / 100 g = 0.011 0.011 * 87.2 g =0.9592 =0.96 g Exercise 56 in Ch. 14 A 5.00-mL sample of an H 3 PO 4 solution of unknown concentration is titrated with a 0.1003 M NaOH solution. A volume of 6.55 mL of the NaOH solution was required to reach the endpoint. What is the concentration of the unknown H 3 PO 4 solution? Answer: 3NaOH(aq) + H3PO4(aq) --> Na3PO4(aq) + 3H2O(l) 3 moles NaOH are required to neutralise 1 mole of H3PO4 moles NaOH used = molarity x litres = 0.1003 M x 0.00655 L = 0.000656965 moles NaOH 3 moles NaOH neutralise 1 mole H3PO4, then Unformatted text preview: moles H3PO4 = 1/3 x moles NaOH = 0.000218988 moles This is how many moles of H3PO4 were in the 5.00 ml sample titrated Molarity H3PO4 = moles / litres = 0.000218988 mol / 0.00500 L = 0.0438 M Exercise 62 in Ch. 15 An equilibrium mixture of the following reaction was found to have [SO 3 ] = 0.391 M and [O 2 ] = 0.125 M at 600 °C. What is the concentration of SO 2 ? 2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g ) K eq = 4.34 at 600 °C Answer: As Keq = [SO3]^2 / [O2] / [SO2]^2 4.34=[(0.391^2)/0.125]*[SO2]^2 4.34*0.125=(0.391^2)*[SO2]^2 (0.391^2) / [4.34*0.125]=[SO2]^2 0.152881 / 0.5425 =SO2^2 0.28180 = SO2^2 SO2=0.531 M Exercise 52 in Ch. 16 What is the oxidation state of Cr in each of the following compounds? A. CrO = +2 B. CrO 3 = +6 C. Cr 2 O 3 = +3...
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• Winter '12
• EarlGurley
• Chemistry, Mole, Sodium hydroxide, Molar concentration

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