PracticeSheet8 key (1)

PracticeSheet8 key (1) - Practice Sheet 8 BLC DLC answers...

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Practice Sheet 8 BLC & DLC answers 8B.1 Remember to do the entire calculation with more than the required number of significant figures so as to avoid round-off errors. Then, last of all, round the final answer to 3 significant figures as the problem requests. a) HClO 4 = H + Cl + 4O = 1.008 + 35.453 + (4 × 15.9994) = 100.459 ≈ 100. g/mol b) NaOH = Na + O + H = 22.990 + 15.999 + 1.008 = 39.997 ≈ 40.0 g/mol c) Mg(NO 3 ) 2 = Mg + 2N + 6O = 24.305 + (2 × 14.007) + (6 × 15.999) = 148.313 ≈ 148 g/mol d) Xenon trioxide XeO 3 = Xe + 3O = 131.293 + (3 × 15.999) = 179.290 ≈ 179 g/mol 8B.2 a) 2P + 3Cl 2 = 2PCl 3 (phosphorous trichloride) P + 3Cl = 30.974 + (3 × 35.453) = 137.33 g/mol → MW; this is a molecular compound. (Note: in excess Cl 2 , this will go on to make PCl 5 , but we’ll stop at the trichloride) b) 3O 2 + 4Mn = 2Mn 2 O 3 (manganese (III) oxide) 2Mn + 3O = (2 × 54.938) + (3 × 15.999) = 157.87 g/mol → FW; it’s an ionic compound. c) 3S + 2Al = Al 2 S 3 (aluminum sulfide) 2Al + 3S = (2 x 26.982) + (3 x 32.065) = 150.139 g/mol FW ionic compound d) N 2 + 3H 2 = 2NH 3 (ammonia) N + 3H = 14.007 + (3 × 1.008) = 17.03 g/mol → MW; ammonia is molecular.
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