A 1.0 g sample of an alkaline earth

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 + Cl2 = MCl2 1 mol of M reacts with 1 mol Cl2 0.8092g/71g/mol = 0.0114 mol 1 g M/gAt Wt M = 0.0114 mol M = 87.7 g/mol A check of the periodic table reveals that M is? If 9.46 kJ/mol of heat was released, how much heat would be released if 1 mol of SrCl2 was formed? -9.46 kJ/g *87.7 g/mol = -829.6 kJ/mol Sr Sodium nitrite, NaNO2, 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 I3- and the titrating the I3- liberated with thiosulfate (S2O3-2) to form iodide ion and S4O6-2. When a nitrite containing sample of meat (2.935g) was analyzed, 18.77 mL of 0.15 M Na2S2O3 solution was needed for the analysis. What is the mass percentage of sodium nitrite in the meat sample? NaNO2 + I- = NO (g) + Na + I3- oxidation state of N in NaNO2? oxidation state of I-? -1 +3 In I3- In NO? +2

-1/3 e- +NaNO2 + 2H+ = NO(g) + H2O + Na+ 3I- = I3- + 2e- reduction oxidation 2NaNO2 + 4H+ +3I- = 2NO(g) + 2H2O + I3- + 2 Na+ 2NaNO2 + 4H+ +3I- = 2NO(g) + 2H2O + I3- + 2Na+ Titrating the I3- liberated with thiosulfate (S2O3-2) forms iodide ion and S4O6-2. oxidation state of I3-? In I- ? -1/3 oxidation state of S in S2O3-2?

+2 I3- +2 e- = 3I2S2O3-2 = S4O6-2 +2 e- 2S2O3-2 + I3- = S4O6-2 + 3I- -1 In S4O6-2 ? +2.5 2NaNO2 + 4H+ +3I- = 2NO(g) + 2H2O + I3- + 2Na+ 2S2O3-2 + I3- = S4O6-2 + 3IWhen a nitrite containing sample of meat (2.935g) was analyzed, 18.77 mL of 0.15 M Na2S2O3 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 Na2S2O3 How many mol of I3- must have been present?

0.0014 mol I3- Therefore 0.0014 mol of I3- must have been produced in the first reaction How many mol of NaNO2 are needed to produce 0.0014 mol I3-? 0.0028 mol NaNO2 MW NaNO2 = 69 g/mol 0.0028 mol NaNO2* 69 g/mol = 0.193g 0.193/2.935*100 = 6.6 % 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*1023 atoms/mol = 51.5*10-23 = 51.5*10-20 J/atom

E = h*; 51.5*10-20 J/atom= 6.63*10-34 Js* = 7.77*1014 s-1 3*108 cm/s = *7.77*1014 s-1 = 386 nm 3*108 m/s = * = 3.86*10-7 m *109 nanometers/m 386 nm Three atoms have the following electronic configurations: a) 1s22s22p63s23p1 Al b) 1s22s22p63s23p5 Cl c) 1s22s22p63s23p64s1

K Which of the three has the largest Ei1? Cl Which of the three has the smallest Ei4? Cl A 0.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M2O) and nitride (M3N). 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 + O2 = M2O Rx 2 M+ N2 = M3N Rx 1: M2O + H2O = MOH HCl + MOH + NH3 = What is the first thing to do? Rx 1: 4M + O2 = 2M2O

1/2M2O + 1/2H2O = MOH Rx 2: 6M+ N2 = 2M3N Rx 2: M3N + H2O = MOH + NH3 MCl + H2O + NH4Cl balance the equations! M + 1/4O2 = 1/2M2O MOH + HCl = MCl + H2O M + 1/6N2 = 1/3 M3N 1/3M3N + H2O = MOH + 1/3NH3 MOH +1/3NH3 + 4/3HCl = MCl + 1/3NH4Cl +H2O A 0.053 g sample of an alkali metal was burned in air to give a mixture of oxide (M2O) and nitride (M3N). 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 Amount of HCl consumed to endpoint = 0.0968 L*0.1M; 0.00968mol How much HCl would be required if M is Li: 0.053/6.9 =0.00764 mol Li only Rx 1 occurs: 0.0764*1= 0.00764 mol only Rx 2 occurs: 0.0764*4/3= 0.01019 mol How much HCl would be required if M is Na: 0.053/23 =0.0023 mol Na only Rx 1 occurs: 0.0023 *1= 0.0023 mol only Rx 2 occurs: 0.0023 *4/3= 0.0031 mol A 0.053 g sample of an alkali metal was burned in air to give a mixture

of oxide (M2O) and nitride (M3N). 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 compared to 0.00968mol 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

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