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DATA AND ANALYSIS Making up the 0.20 M (NH4)2S2O8 solution Show the calculations for the moles and mass of (NH4)2S2O8 that is required for 60 mL of a 0.20 M (NH4)2S2O8 solution. (NH4)2S2O8 = 228.18 g/mol 1 mole of a substance in 100 ml solvent = 1 M solution 0.2 mole of a = 0.2 M solution The weight of the 0.2 mole of (NH4)2S2O8 = (228.18 x 0.2) = 45.636 g 0.2 (M) 100 mL (NH4)2S2O8 solution contains 65.636 g solute 0.2 (M) 60 mL (65.636 x (60) / (1000)) = 2.738 g To make a 60 mL 0.20 M (NH4)2S2O8 solution one needs 2.738 g of (NH4)2S2O8 solute Clock reaction data and calculations The data that you put into Table 3 will be used to compute the rates of each of the three clock reactions. The moles of I2 produced are (1) / (2) the number of moles of thiosulfate added (see discussion on page 2:4). The total volume in the flask is given in Table 2, but you must convert this into units of liters. With this data, solve for molarity of I2 and for the rate of I2 formation. Notes: the values in each of the three middle columns are the same. Consider using the same exponent for the rates to make comparing them easier. Table 3 I2 Produced Time (s) (mole) Total Volume (L) Molarity of I2 [l] Rate 2:9 -- 

          DATA AND ANALYSIS
Making up the 0.20 M (NH4)2S2O8 solution
Show the calculations for the moles and mass of (NH4)2S2O8 that is required for 60 mL of a 0.20 M (NH4)2S2O8 solution.
(NH4)2S2O8 = 228.18 g/mol
1 mole of a substance in 100 ml solvent = 1 M solution
0.2 mole of a = 0.2 M solution
The weight of the 0.2 mole of (NH4)2S2O8 = (228.18 x 0.2) = 45.636 g
0.2 (M) 100 mL (NH4)2S2O8 solution contains 65.636 g solute
0.2 (M) 60 mL (65.636 x (60) / (1000)) = 2.738 g
To make a 60 mL 0.20 M (NH4)2S2O8 solution one needs 2.738 g of (NH4)2S2O8 solute
Clock reaction data and calculations
The data that you put into Table 3 will be used to compute the rates of each of the three clock reactions.
The moles of I2 produced are (1) / (2) the number of moles of thiosulfate added (see discussion on page 2:4). The total volume in the flask is given in Table 2, but you must convert this into units of liters. With this data, solve for molarity of I2 and for the rate of I2 formation. Notes: the values in each of the three middle columns are the same. Consider using the same exponent for the rates to make comparing them easier.
Table 3
I2 Produced Time (s) (mole)
Total Volume (L)
Molarity of I2 [l]
Rate
2:9
--
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data and analysis making up the 020 m nh42s2o8 solution show the calculations for the moles and mass of nh42s2o8 that is required for 60 ml of a 020 m nh42s2o8 solution nh42s2o8 22818 gmol 1 10096

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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DATA AND ANALYSIS Making up the 0.20 M (NH4)2S2O8 solution Show the calculations for the moles and mass of (NH4)2S2O8 that is required for 60 mL of a 0.20 M (NH4)2S2O8 solution. (NH4)2S2O8 = 228.18 g/mol 1 mole of a substance in 100 ml solvent = 1 M solution 0.2 mole of a = 0.2 M solution The weight of the 0.2 mole of (NH4)2S2O8 = (228.18 x 0.2) = 45.636 g 0.2 (M) 100 mL (NH4)2S2O8 solution contains 65.636 g solute 0.2 (M) 60 mL (65.636 x (60) / (1000)) = 2.738 g To make a 60 mL 0.20 M (NH4)2S2O8 solution one needs 2.738 g of (NH4)2S2O8 solute Clock reaction data and calculations The data that you put into Table 3 will be used to compute the rates of each of the three clock reactions. The moles of I2 produced are (1) / (2) the number of moles of thiosulfate added (see discussion on page 2:4). The total volume in the flask is given in Table 2, but you must convert this into units of liters. With this data, solve for molarity of I2 and for the rate of I2 formation. Notes: the values in each of the three middle columns are the same. Consider using the same exponent for the rates to make comparing them easier. Table 3 I2 Produced Time (s) (mole) Total Volume (L) Molarity of I2 [l] Rate 2:9 --
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Sri K.
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Transcript

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00:01 In this question we are asked to write down the reaction when sulfur dioxide reduces aqua solution of potassium nitrate kmno4 potassium per manganate to manganese sulfate which is also in aqua state.
00:23 So we can write the balanced chemical equation as mno4 negative is what we will use for kmnmno4.
00:34 Then h plus we will be using sulfuric acid as it is an acetic medium mn2 plus belongs to the compound mnso4 and k plus is the ketion of k2so4 so we can write the reaction in this manner twice of kmno4 in at first state when added to five oxygen molecules it is actually 5 s .o2, 5 s .o2 in aqua state.
01:14 Added to two water molecules...
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