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1. 19 F: 9e –, 9p +, 10n 0 Ar-40: 18e –, 18p +, 22n 0 2.a) Al 2 S 3 b) MgO 3.NaF = 4.0 – 0.9 = 3.1: ionic, London. H 2 O = 3.5 – 2.1 = 1.4: dipole-dipole.

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Presentation on theme: "1. 19 F: 9e –, 9p +, 10n 0 Ar-40: 18e –, 18p +, 22n 0 2.a) Al 2 S 3 b) MgO 3.NaF = 4.0 – 0.9 = 3.1: ionic, London. H 2 O = 3.5 – 2.1 = 1.4: dipole-dipole."— Presentation transcript:

1 1. 19 F: 9e –, 9p +, 10n 0 Ar-40: 18e –, 18p +, 22n 0 2.a) Al 2 S 3 b) MgO 3.NaF = 4.0 – 0.9 = 3.1: ionic, London. H 2 O = 3.5 – 2.1 = 1.4: dipole-dipole (specifically H-bonding), London. O 2 = 3.5 – 3.5 = 0: London 4.H: 3.03/98 x 100= 3.1% H P: 30.97/98 x 100= 31.6% P O: 64/98 x 100= 65.3% O x 2.78 = = = (0.199)(10) + (0.801)(11) = mol C 2 H 4 x 6.02x10 23 particles 1 mol C 2 H 4 = 2.64x10 25 atoms x 6 atoms 1 particle C 2 H 4

2 8.# g CuSO 4 = 3.8 mol x g/mol = 606 g g C x 1 mol/12.01 g = 5.26 mol (3.01) 8.8 g H x 1 mol/1.01 g = 8.71 mol(4.98) 28.0 g O x 1 mol/16.0 g = 1.75 mol (1) Simplest formula = C 3 H 5 O 10.CH 2 O: 30 g/mol, 180 g/mol 30 g/mol = 6 Molecular formula: C 6 H 12 O Bi 4 2 He Tl Se 0 –1 e Br 12.# mL O 2 = 1 mL C 4 H 10 x 13 mL O 2 /2 mL C 4 H 10 = 6.5 mL used, 2.5 mL remain 13. # g O 2 = 592 g KNO 3 1 mol KNO g KNO 3 = 93.7 g O 2 1 mol O 2 2 mol KNO 3 xx g O 2 1 mol O 2 x

3 14. 1 mol Fe 3 O 4 3 mol Fe x 4.91 mol Fe 379 g Fe 3 O 4 = g Fe 3 O 4 1 mol Fe 3 O 4 x 2 mol O 2 x # g Fe 3 O 4 = 100 g O g Fe 3 O 4 = g Fe 3 O 4 1 mol Fe 3 O 4 x 1 mol O g O 2 x mol BaSO 4 1 mol BaCl 2 x # g BaSO 4 = 18.0 g BaCl g BaSO 4 = g BaSO 4 1 mol BaSO 4 x 1 mol BaCl g BaCl 2 x % yield = actual theoretical x 100% = 13.2 g g x 100% = 65.4% g 42 g x 100% = 35.7%

4 17. ppm = mg/kg, 4.0 ppm = x mg / kg x = mg = 200 g = 0.20 kg 18. # mol = 16 g x 1 mol / 40 g = 0.40 mol NaOH [NaOH] = 0.40 mol / 2.00 L = 0.20 M 19.M 1 V 1 = M 2 V 2 (12.0 M)(V 1 ) = (0.175 M)(3.00 L) V 1 = L = 43.8 mL L x 1.3 mol/L L x 0.75 mol/L = 6.05 mol 3.5 L L = 5.5 L mol/L = 6.05 mol / 5.5 L = 1.1 mol/L 21.g/100 mL H 2 O: at 52 C = 90, at 20 C = – 32 = 58 g/100 mL H 2 O precipitates 58 g/100 mL x 70 mL = 40.6 g precipitate

5 22. First, we need a balanced equation: 2Na 3 PO 4 (aq) + 3CaCl 2 (aq) Ca 3 (PO 4 ) 2 (s)+ 6NaCl(aq) 1 mol Ca 3 (PO 4 ) 2 3 mol CaCl 2 x # g Ca 3 (PO 4 ) 2 = 0.35 L 6.3 g Ca 3 (PO 4 ) 2 = 310 g Ca 3 (PO 4 ) 2 1 mol Ca 3 (PO 4 ) 2 x mol CaCl 2 1 L x 23.[H + ] = 10 –pH = 10 –7.5 = 3.16 x 10 –8 pH = – log [H + ] = – log [1.85x10 –12 ] = M A x V A x #H = M B x V B x #OH (0.600 mol/L)(4.42 mL)(2) = (M B )(5.00 mL)(1) M B = 1.06 mol/L 25.K = °C – 273, or °C = K a) 22 K = -251°C, b) 756°C = 1029 K

6 P 1 = 150 kPa, T 1 = 298 K P 2 = ? kPa, T 2 = 473 K, V 1 =V 2 P 2 = 238 kPa P1P1 T1T1 = P2P2 T2T2 26. P1V1P1V1 T1T1 = P2V2P2V2 T2T2 27. P 1 = 103 kPa, V 1 = 22 L, T 1 = 288 K P 2 = 15 kPa, V 2 = 108, T 2 = ? T 2 = 206 K = –67 C 28. P 1 = 97 kPa – 2.34 kPa = kPa V 1 = 35.7 mL, T 1 = 293 K P 2 = kPa, V 2 = ?, T 2 = 273 K V 2 = 31 mL 29.PV = nRT (8.31 kPaL/Kmol)(295 K) (110 kPa)(1.46 L) = mol n = g/mol = 2.62 g/ mol = 40.0 g/mol

7 30. 2 mol H 2 O 1 mol O 2 x # g H 2 O= 2.4 L O g H 2 O= g H 2 O 1 mol H 2 O x 1 mol O L O 2 x 31.C 27 H O 2 27 CO H 2 O 32.# mol = 1.37 g g/mol = mol q = cm T = (4.18 J/g C)(2300 g)(2.37 C) = J J/mol= J mol = 5694 kJ/mol

8 H-Br Br-Br H-H released#required#kJ/molBond H 2 + Br 2 2HBr kJ H 2 + Br 2 2HBr For more lessons, visit


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