Exam #1 - Chemistry
1. Nitrogen monoxide (NO) is the smallest hormone known, playing a role in numerous processes in the nervous, immune and cardiovascular systems. Use the phase diagram of nitrogen monoxide to answer the following. Show work. a. Indicate and name the phase change(s) that take place when NO is cooled from –160 oC to –165 oC at a constant pressure of 250 torr. (3 pts.) gas liquid condensation b. freezing Indicate the phase change(s) that take place when the pressure on NO is increased from 100 to 1000 torr at a constant temperature of –164 oC. (3 pts.) gas solid deposition c. solid liquid melting Estimate the normal boiling point of NO. (2 pts.) at 760 torr, –156.8 ºC d. Estimate the triple point of NO. (2 pts.) 190 torr, –163.5 ºC f. Use the phase diagram to explain whether solid NO would float or sink in liquid NO at its freezing point? (2 pts.) The melting point line slopes to the left. This means that the most dense phase is liquid NO. Solid NO would float (like ice in water). 2. Consider the following pure substances: (28 pts.) BCl3 SiO2(quartz) SrCl2 HClO4 a. Classify each as molecular, ionic or other. molecular other ionic molecular Xe POBr3 Cr2O3 ICl3 other molecular ionic molecular b. Give the valence electron count and the Lewis structure of each substance you classified as molecular. Do not show shape. For ionic compounds, give charges on the cation and anion. BCl3 SiO2 24 ve– SrCl2 Sr2+ Cl– HClO4 Cl Cl B Xe 32 ve– O POBr3 O Cl ICl3 Cr3+ O2– 28 ve– O H Cl Br Cl Cr2O3 32 ve– P I Cl Br Cl O Br O c. Draw the VSEPR shape of each substance you classified as molecular (without lone-pair e– or multiple bonds). Clearly show, using vector notation on the shape, whether the molecular compounds are polar or nonpolar. BCl3 SiO2 SrCl2 HClO4 Xe Cl POBr3 H O P B Cl Cl Cl O nonpolar polar Cr2O3 ICl3 O Br O O Cl I Br Br Cl polar polar d. Name the dominant intermolecular force in each substance. BCl3 SiO2 London 56 e– covalent bond SrCl2 HClO4 ion-ion H-bonding Xe POBr3 London 54 e– 2x1 Cr2O3 dipoledipole 128 e– ICl3 ion-ion 3x2 dipoledipole 104 e– BCl3 Xe e. List the substances in order of decreasing (highest to the left) boiling point. SiO2 Cr2O3 SrCl2 HClO4 POBr3 ICl3 f. Give a specific, quantitative reason for the order if the intermolecular forces are the same. Product of charges on the ions (for ionic); total electrons for dipole-dipole and London. See electron totals on d. Cl 3. Classify the type of solid formed by each of the substances in question 2 and the substance in question 4. (5 pts.) BCl3 POBr3 molecular SiO2(quartz) covalent (network) Cr2O3 ionic SrCl2 ionic ICl3 molecular HClO4 molecular Cs metallic Xe atomic Cesium is so reactive, not many people have seen a pure sample. Its melting point is so low it is close to being only the third liquid element on the Periodic Table (after bromine and mercury). Use the following data to calculate the heat needed to take 265.8 g of cesium from 25.0 oC to 640.0 oC. (10 pts.) 28.4 oC 641.0 oC 2.092 kJ/mole 67.74 kJ/mole 0.242 J/g∙oC 0.244 J/g∙oC 0.156 J/g∙oC normal melting point Cs(s) normal boiling point Cs(ℓ) ΔHofusion Cs(s) ΔHovaporization Cs(ℓ) specific heat capacity Cs(s) specific heat capacity Cs(ℓ) specific heat capacity Cs(g) 95 85 Temperature (oC) 4. molecular 75 65 55 45 35 25 0 1000 2000 3000 4000 Heat (J) 5000 6000 7000 8000 (part of heating curve) 1. q = msT = 265.8 g x 0.242 J/g∙oC x (28.4 – 25.0 oC) = 218.70 J 2. Hfusion = 265.8 g x 1 mole/132.905 g x 2092 J/mole = 4183.84 J 3. q = msT = 265.8 g x 0.244 J/g∙oC x (640.0 – 28.4 oC) = 39665.44 J qtotal = 218.70 J + 4183.84 J + 39665.44 J = 44.1 kJ 5. Copper is one of the few elements that can be found in nature as the free element. Copper crystallizes in a body-centered-cubic lattice (bcc) and has a density of 8.96 g/cm3. Calculate the atomic radius of copper. (10 pts.) 63.546 g Cu 1 mole Cu 2 atoms Cu 1 cm3 x x x 2.3554 x 1023cm3 23 1 mole Cu 6.022 x 10 atoms Cu 1 unit cell 8.96 g Cu a 3 V 3 2.3554 x 1023cm3 2.8665 x 108cm a 3 2.8665 x 108 cm 3 r 1.24 x 10–8 cm = 1.24 Å = 124 pm 4 4 6. Iron(III) sulfate has been detected by the two martian rovers Spirit and Opportunity and is indicative of strongly oxidizing conditions of the surface of Mars. In fact, the Spirit rover became stuck when it drove over a patch of soft iron(III) sulfate that had been hidden under some normal soil. This meant the end of the mission. A solution is prepared by dissolving 144.0 g of iron(III) sulfate (FW 399.89 g/mole) in 360.0 mL of water (FW 18.015 g/mole). Calculate: a. The molality of the sample. (5 pts.) 1 mole Fe 2 (SO 4 )3 399.89 g Fe 2 (SO 4 )3 1 g H 2O 1 kg 360.0 mL H 2O x x 1 mL H 2 O 1000 g 144.0 g Fe2 (SO 4 )3 x m = 1.000 m b. The normal freezing point of the solution. (5 pts.) Tf iKf m (5)(1.86 o C/m)(1.000 m) = 9.30 oC lower Tf = 0.0 oC – 9.30 oC = –9.30 oC c. The vapor pressure of the solution at 30.0 oC. (5 pts.) χ water 1 mole H 2O 18.015 g H 2O = 0.9173 19.9833 mole H 2O 5 0.3601 mole Fe2 (SO4 )3 Psolution 360.0 g H 2 O x χ water Po water = 0.9173 x 31.82 torr = 29.19 torr (0.03841 atm) 7. Pupfish are a small species of fish noted for being able to live in extreme conditions. They are found in the very saline, very hot Salt Creek in Death Valley National Park. Because of the temperature, the amount of oxygen in the water is reduced, but that is partially offset because of the higher atmospheric pressure below sea level. Fish generally need an O 2 concentration in water of at least 4 mg/L for survival. Use the data on the Periodic Table page to determine the concentration of O 2 (in mg/L) in Salt Creek at 44.4 ºC (112 ºF) and an air pressure of 774.4 torr. (10 pts.) 1 atm = 0.2134 atm PO2 = χ O2 Ptotal = 0.209476 x 774.4 torr x 760 torr CO2 k O2 PO2 9.331 x 104 M/atm O2 x 0.2134 atm O2 = 1.989 x 104 M O2 1.989 x 104 mole O2 31.998 g O2 1000 mg O2 = 6.363 mg O2/L x x L 1 mole O2 1 g O2
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