Solutions to the Extra Problems for Chapter The temperature is K. To convert to Kelvin, we use Equation (10.4):

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Juniper Stewart
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1 Solutions to the Extr roblems for Chpter 0. he temperture is 98. K. o convert to Kelvin, we use Eqution (0.4): K ºC Since we re dding, the significnt figures re determined by precision. 5.0 is the lest precise number, becuse its lst significnt figure is in the tenths plce. hus, the nswer must be to the tenths plce s well.. he temperture is -78 C. o convert between Kelvin nd Celsius, we use Eqution (0.4): K ºC ºC ºC C Since 95 hs its lst significnt figure in the ones plce, the nswer must hve its lst significnt digit in the ones plce. 3. he volume is 86.5 ml. Since the temperture doesn t chnge, nd re the sme. ht mens they cncel in the combined gs lw. Since 567 torr nd 49.6 ml go together, we will cll them nd V. ht mens 98 torr is. (567 torr) (49.6 ml) (98 torr) V Dividing by 98 torr solves for V. (567 torr) (49.6 ml) V 86.5 ml 98 torr 4. he volume is 3.7 L. Since pressure doesn t chnge, nd cncel out in the combined gs lw. he volume of 6.7 liters goes with 5.0 C, so those re V nd. his mkes 50.0 C. However, we need to convert both those tempertures to Kelvin first. K ºC K ºC

2 Multiplying by 43. K solves for V : 6.7 L V 98. K 43. K 6.7 L V ( L 98. K 5. he volume is 40 liters. In this problem, nothing is held constnt. he volume of 5.6 liters goes with 5.0 C nd.0 tm, so those re V,, nd. ht mens -. C is nd 0.0 tm is. Before we use the combined gs lw, however, the tempertures must be in Kelvin. K ºC K ºC (.0 tm) (5.6 L) (0.0 tm) 98. K 6. K Multiplying by 6. K nd dividing by 0.0 tm solves for V : (.0 tm) (5.6 L) (6. V 40 L 98. K (0.0 tm) 6. here re moles of wter vpor. he Idel Gs Lw reltes moles to pressure, volume, nd temperture, but the units need to be tm, liters, nd K: 745 torr 65 ml tm 760 torr 0.00 L ml tm 0.65 L K ºC Now we cn use the Idel Gs Lw: V nr L tm (0.980 tm) (0.65 L) n 0.08 (94.7

3 o solve for n, we need to divide both sides by 0.08 (L tm)/(mole nd 94.7 K: (0.980 tm) (0.65 L) n moles L tm 0.08 ( he volume is 5 L. here is no reltionship mong grms, volume, pressure, nd temperture for gs, but the Idel Gs Lw reltes moles, volume, pressure, nd temperture, s long s pressure is in tm nd temperture is in K. hus, we must get from grms to moles nd convert the pressure nd temperture to the right units g NH mole NH moles NH g NH3 869 torr tm 760 torr.4 tm K ºC V nr Dividing by.4 tm solves for V: L tm (.4 tm) (V) (5.869 moles) 0.08 (357.7 mole K L tm (5.869 moles) 0.08 (357.7 mole K V 5 L.4 tm 8. here re.5 L of hydrogen gs produced. If we wnt to get the volume of hydrogen, we will need to figure out the moles of hydrogen. We cn get tht by doing stoichiometry. 5.0 g Mg mole Mg 4.3 g Mg.03 moles Mg Now we cn use the chemicl eqution to convert to O :.03 moles Mg mole H.03 moles mole Mg H We now hve moles for the Idel Gs Lw, but we need to convert the temperture to Kelvin:

4 K ºC V nr Dividing by. tm solves for V: L tm (. tm) (V) (.03 moles) 0.08 (98. mole K L tm (.03 moles) 0.08 (98. mole K V.5 L. tm 9. She will need 9 L. We could use the Idel Gs Lw, but since the things we re interested in re both gses, we cn just use Avogdro s Lw nd llow the chemicl eqution to give us the reltionship between the volumes. 6 liters NH 3 liters H 3 9 liters H liters NH3 0. he mole frction of hydrogen is 0.675, nd the mole frction for nitrogen is he mole frction is relted to the prtil pressure nd totl pressure. We hve the prtil pressures, so we just hve to get the totl pressure: totl 75 torr torr torr X totl 75 torr torr Xhydrogen torr torr Xnitrogen he prtil pressures re.4 tm for rgon,.4 tm for hydrogen, nd. tm for nitrogen. Eqution (0.9) llows us to figure out the prtil pressure of ech gs if we know the mole frction. ht s our first step. moles of component X totl moles in the mixture otl moles in the mixture 0.54 moles moles moles.9 moles

5 0.54 moles X rgon moles 0.9 moles X hydrogen moles 0.45 moles X nitrogen moles We cn now get the prtil pressures from the totl pressure: X totl X totl rgon (0.8) (5. tm).4 tm hydrogen (0.48) (5. tm).4 tm nitrogen (0.4) (5. tm). tm. he chemist mde 0.05 moles. We cn t use the Idel Gs Lw immeditely, becuse the gs is collected over wter. hus, it is mixture of hydrogen nd wter vpor. o subtrct out the wter vpor, we need the vpor pressure of wter t 30.0 C. he tble on pge 3 sys it is 3.8 torr. However, to use the Idel Gs Lw, tht hs to be in tm. 3.8 torr tm 760 torr tm he pressure given is the totl pressure. According to Dlton s Lw of rtil ressure: totl + b 0.9 tm tm + oxygen oxygen 0.9 tm tm tm o use the Idel Gs Lw, temperture must be in Kelvin. K ºC C V nr

6 L tm (0.879 tm) (.45 L) n 0.08 (303. o solve for n, we need to divide both sides by 0.08 (L tm)/(mole nd 303. K: (0.879 tm) (.45 L) n 0.05 moles L tm 0.08 (303.

[ HOCl] Chapter 16. Problem. Equilibria in Solutions of Weak Acids. Equilibria in Solutions of Weak Acids

[ HOCl] Chapter 16. Problem. Equilibria in Solutions of Weak Acids. Equilibria in Solutions of Weak Acids Equilibri in Solutions of Wek Acids Chpter 16 Acid-Bse Equilibri Dr. Peter Wrburton peterw@mun.c http://www.chem.mun.c/zcourses/1011.php The dissocition of wek cid is n equilibrium sitution with n equilibrium