Question

254 CHAPTER 7 Chemical Quantities and Reactions 7.81 A dandruff shampoo contains dipyrithione. C N.OS which acts as an antiba
U 21 20 93 How many grams are in 0.150 mole of each of the following? (7.3) b. Cl2 c. Na2CO3 a. K
793 How many grams are in 0.150 mole of each of the following? (7.3) a. K b. Cl2 c. Na2CO3
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Answer #1

7.810) . a). Molecular mass of dipyrithione (C10H8N2O2S2) = 12x10 + 8 + 14x2 + 16x2 + 32x2 = 252 g/mol

b). mole in 25 g of it = 25 / 252 = 0.0992 mol

c). 1 mol of dipyrithione have 10 mole of C atom

then , 0.0992 mol of dipyrithione have 10 x 0.0992 mole = 0.992 mol of carbon

d). mole of N in 8.2 x 10^24 = 8.2x10^24 / (6.022x10^23) = 13.616 mole

2 mol of nitrogen is present in 1 mole of dipyrithione (c10h8n2o2s2 )

1mol of nitrogen is present in 1/2 mole of dipyrithione

so, 13.616 mol of nitrogen is present in 13.616 / 2 mole of dipyrithione = 6.808 moles .

7.93. a). grams in 0.15 mole of k = 0.15 x mol wt. of K

= 0.15 x 39 = 5.85 g

b) . grams of Cl2 = 0.15 x 70.9 = 10.635 g

c). grams of Na2CO3 = 0.15 x 286 = 42.9 g

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