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chem3-solutions

# chem3-solutions - Solutions Chem 1201 Final Exam Fall 2005...

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Solutions ————————————————————————————————— Chem 1201 Final Exam Fall 2005 ————————————————————————————————————— 1 Solution: 2 Solution: 0.1990 10.0129 amu 0.8010 11.0093 amu 10.811 amu/atom 3 Solution: In HCN and H 2 CO the H atom is not bound to an N,O, or F. This is also true for HBr and BeH 2

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4 Solution: Angle 1 has a central N atom with 4 domains. This is a tetrahedral domain geometry with 109.5 0 bond angles. The angle is a little less than 109.5 0 because of the nonbonding electrons which tend to repel more than bonding domains. The problem doesn’t have any choices like that. The closest choice is 109.5 0 We will go with that. Angle 2 has a central C atom with 3 domains. This is a trigonal plane domain geometry with 120 0 bond angles. 5 Solution: I always begin a problem by asking what I need to answer what the question asks. To find molarity I will use the equation M moles H 2 O 2 vol I need the moles of H 2 O 2 . How will I find it? We are given volume and concentration of MnO 4 . Let’s calculate the moles of MnO 4 used in this reaction. vol conc moles 0.0152 L 0.103 M 0.00157 moles MnO 4 But how are moles MnO 4 related to moles H 2 O 2 we need? That is what the chemical recipe tells us. Here is the relationship from the recipe 5 mol H 2 O 2 2 mol MnO 4 So moles of H 2 O 2 is 0.00157 moles MnO 4 5 mol H 2 O 2 2 mol MnO 4 0.00391 moles H 2 O 2 Concentration of H 2 O 2 is M moles H 2 O 2 vol 0.00391 moles H 2 O 2 0.01 L 0.391 M This problem is really a titration problem. 6
Solution: u 3 RT M Before using this equation remember the common mistakes !!! Mass should be in kg and R 8.314 J/mol K, T in Kelvins! u 3 8.314 J/mol K 300 K 352.03 10 3 kg 145.8 m/s 7 Solution: V 1 T 1 V 2 T 2 4.00 L 273.15 27 V 2 273.15 10 V 2 3.8 L 8 Solution: atomic number number of protons 35 mass number protons neutrons 35 45 80 electrons determine the charge: There are 35 protons and 36 electrons so the charge is minus 1. (Another way to think about this is there is one extra electron which makes the charge 1.) Look up element with atomic number 35 in the periodic table. Its Br 35 80 Br 9 Solution: Assume 100g. That means we have a sample with 38.8 g Ca, 20.0 g P, 41.3 g O. Now convert to moles. 38.3 g Ca 40.08 g/mol 0.9555 mol 20.0 g P 30.97 g/mol 0.6458 mol 41.3 g O 16.00 g/mol 2.5813 mol The ratio of atoms in the empirical formula is the same as the mole ratios

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Ca 0.9555 P 0.6458 O 2.5813 Divide all subscripts by the smallest to get the simplest whole number ratio. Ca 1.480 P 1 O 3.997 Do not round by more than 0.1. We can round the answer to Ca 1.5 P 1 O 4 To get whole numbers we have to multiply all subscripts by 2 Ca 3 P 2 O 8 10 Solution: Effective nuclear charge simply means the total charges (from protons) experienced by electrons. The words in this problem " effective nuclear charge is larger for Si " simply mean that Si has more protons than A
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