1. A 1.0 g sample of an alkaline earth metal M reacts completely with 0.8092 g of chlorine gas to yield and ionic salt with the formula MCl 2. In the process 9.46 kJ of heat is released. What is the molecular mass and identity of the metal M? M + Cl 2 = MCl 2 1 mol of M reacts with 1 mol Cl 2 0.8092g/71g/mol = 0.0114 mol 1 g M/gAt Wt M = 0.0114 molM = 87.7 g/mol A check of the periodic table reveals that M is?Sr If 9.46 kJ/mol of heat was released, how much heat would be released if 1 mol of SrCl 2 was formed? -9.46 kJ/g *87.7 g/mol = -829.6 kJ/mol

2. Sodium nitrite, NaNO 2, is frequently added to processed meats as a preservative. The amount of nitrite in a sample can be determined by reducing the nitrite to nitric oxide (NO) in acid with excess iodide which forms I 3 - and the titrating the I 3 - liberated with thiosulfate (S 2 O 3 -2 ) to form iodide ion and S 4 O 6 -2. When a nitrite containing sample of meat (2.935g) was analyzed, 18.77 mL of 0.15 M Na 2 S 2 O 3 solution was needed for the analysis. What is the mass percentage of sodium nitrite in the meat sample? NaNO 2 + I - = NO (g) + Na + I 3 - oxidation state of N in NaNO 2 ?+3In NO?+2 oxidation state of I - ?In I 3 - -1/3 e - +NaNO 2 + 2H + = NO(g) + H 2 O + Na + reduction 3I - = I 3 - + 2e - oxidation 2NaNO 2 + 4H + +3I - = 2NO(g) + 2H 2 O + I 3 - + 2 Na +

3. Titrating the I 3 - liberated with thiosulfate (S 2 O 3 -2 ) forms iodide ion and S 4 O 6 -2. oxidation state of I 3 -?-1/3 In I - ? oxidation state of S in S 2 O 3 -2 ?+2In S 4 O 6 -2 ?+2.5 I 3 - +2 e - = 3I - 2S 2 O 3 -2 = S 4 O 6 -2 +2 e - 2S 2 O 3 -2 + I 3 - = S 4 O 6 -2 + 3I -

4. 2NaNO 2 + 4H + +3I - = 2NO(g) + 2H 2 O + I 3 - + 2Na + 2S 2 O 3 -2 + I 3 - = S 4 O 6 -2 + 3I - When a nitrite containing sample of meat (2.935g) was analyzed, 18.77 mL of 0.15 M Na 2 S 2 O 3 solution was needed for the analysis. What is the mass percentage of sodium nitrite in the meat sample? How many mols of sodium thiosulfate was consumed? 0.01877 L * 0.15 M = 0.0028 mol Na 2 S 2 O 3 How many mol of I 3 - must have been present?0.0014 mol I 3 - Therefore 0.0014 mol of I 3 - must have been produced in the first reaction How many mol of NaNO 2 are needed to produce 0.0014 mol I 3 - ? 0.0028 mol NaNO 2 MW NaNO 2 = 69 g/mol 0.0028 mol NaNO 2 * 69 g/mol = 0.193g 0.193/2.935*100 = 6.6 %

5. Photochromic glasses which darken on exposure to light, contain a small amount of silver. When irradiated with light, the silver ion is reduced to silver metal by capture of an electron from chlorine to produce a chlorine atom and the glass darkens. The chlorine atom formed is prevented from diffusing away by the glass. When the light is removed, the silver metal loses an electron and silver chloride is reformed. If 310 kJ/mol of energy is necessary for the reaction to proceed, what wavelength of light is necessary? 310 kJ/mol/6.02*10 23 atoms/mol = 51.5*10 -23 = 51.5*10 -20 J/atom E = h*ν; 51.5*10 -20 J/atom= 6.63*10 -34 Js* ν ν = 7.77*10 14 s -1 3*10 8 m/s = λ*ν 3*10 8 cm/s = λ*7.77*10 14 s -1 λ = 3.86*10 -7 m *10 9 nanometers/m λ = 386 nm

6. 386 nm

7. Three atoms have the following electronic configurations: a)1s 2 2s 2 2p 6 3s 2 3p 1 b)1s 2 2s 2 2p 6 3s 2 3p 5 c)1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Which of the three has the largest E i1 ? Which of the three has the smallest E i4 ? Al Cl K

8. A 0.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M 2 O) and nitride (M 3 N). The reaction product was dissolved in water according to Rx 1 and 2 and titrated with 0.1M HCl requiring 96.8 mL for complete neutralization. What is the metal and composition? Rx 1: M 2 O + H 2 O = MOHRx 2: M 3 N + H 2 O = MOH + NH 3 HCl + MOH + NH 3 = MCl + H 2 O + NH 4 Cl What is the first thing to do?balance the equations! Rx 1: 4M + O 2 = 2M 2 O Rx 2: 6M+ N 2 = 2M 3 N 1/2M 2 O + 1/2H 2 O = MOH 1/3M 3 N + H 2 O = MOH + 1/3NH 3 Rx 1:M + O 2 = M 2 O Rx 2 M+ N 2 = M 3 N M + 1/4O 2 = 1/2M 2 O M + 1/6N 2 = 1/3 M 3 N MOH + HCl = MCl + H 2 O MOH +1/3NH 3 + 4/3HCl = MCl + 1/3NH 4 Cl +H 2 O

9. A 0.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M 2 O) and nitride (M 3 N). The reaction product was dissolved in water and titrated with 0.1M HCl requiring 96.8 mL for complete neutralization according to the following two reactions. Rx (1)with oxygen: 1 mol of M reacts with 1 mol of HCl Rx (2)with nitrogen: 1 mol of M reacts with 4/3 mol of HCl How much HCl would be required if M is Li: only Rx 1 occurs: only Rx 2 occurs: 0.053/6.9 =0.00764 mol Li Amount of HCl consumed to endpoint = 0.0764*1= 0.00764 mol 0.0764*4/3= 0.01019 mol How much HCl would be required if M is Na: only Rx 1 occurs: only Rx 2 occurs: 0.053/23 =0.0023 mol Na 0.0023 *1= 0.0023 mol 0.0023 *4/3= 0.0031 mol 0.0968 L*0.1M; 0.00968mol

10. A 0.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M 2 O) and nitride (M 3 N). The reaction product was dissolved in water and titrated with 0.1M HCl requiring 98.6 mL for complete neutralization according to the following two reactions. Rx (1)with oxygen: 1 mol of M reacts with 1 mol of HCl Rx (2)with nitrogen: 1 mol of M reacts with 4/3 mol of HCl Amount of HCl consumed to endpoint = 0.0986L*0.1; 0.00986 mol How much HCl would it take if Rx 1 and Rx 2 contributed equally? 0.5*.00764+0.5*0.01019 = 0.00891 mol let a be the fraction reacting by Rx 1 a*(0.00764) + (1-a)(0.01019) = 0.00986; a = 0. 148 or 14.8 % by Rx1 compared to 0.00968mol