1307EX2Btest

1307EX2Btest - Equations: o K = oC + 273.15 o F = 1.8 (oC)...

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Equations: o K = o C + 273.15 o F = 1.8 ( o C) + 32 q = m ó c ó T = m ó c ó (T f – T i ) Constants: Specific heat capacity of water: 1.00 cal/g K = 4.184 J/ g K Density of water: 1.00 g/mL Universal gas constant: 0.08206 atm L/mol K = 8.314 J/mol K Conversions: 1 atm = 760 torr (exact) = 760 mm Hg (exact) = 101.325 kPa
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MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1. The percent composition of oxygen in XeO 3 is: A. 8.92% B. 26.8% C. 12.2% D. 36.6% E. 73.2% 2. According to the following equation, if you have 3.5 moles of potassium permanganate, it will react with ____ moles of potassium nitrite: 2 KMnO 4 + 3 KNO 2 2 MnO 2 + 3 KNO 3 + K 2 O A. 2.3 B. 7.0 C. 5.3 D. 3.5 E. 3.0 3. The percentage composition of an organic compound is 84.09 % C and 15.91 % H. What is the empirical formula of this compound? A. CH 2 B. C 7 H 16 C. C 7 H 15 D. C 4 H 9 E. C 4 H 7 4. N 2 H 4 + O 2 N 2 + 2H 2 O The rocket fuel hydrazine, N 2 H 4 , burns cleanly to give products that are not harmful to the environment or the upper atmosphere, water and molecular nitrogen. What mass of water is generated when 64.0 g of N 2 H 4 reacts with 80.0 g of O 2 ? A. 4.0 g B. 18.0 g C. 36.0 g D. 90.0 g E. 72.0 g 5. A particular carbohydrate has the empirical formula CH 2 O and a molar mass of 175 - 185 g/mol. What is the molecular formula of this compound? A. C 6 H 12 O 6 B. C 12 H 24 O 12 C. C 6 H 6 O 6 D. C 2 H 4 O 2 E. C 12 H 6 O 12
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6. What mass of lead is in 15.0 g of PbO 2 ? A. 2.0 g Pb B. 26.0 g Pb C. 9.5 g Pb D. 13.0 g Pb E. 15.0 g Pb 7. Which weighs the most? A. 0.5 mol of oxygen gas from the air B. 1.0 mol of water (MM H2O = 18.02 g/mol) C. 5.0 g of candy D. 6.02 x 10 23 atoms of nitrogen gas E. 0.2 moles of argon 8. When 0.1665 mol of acetic acid (CH 3
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This test prep was uploaded on 04/07/2008 for the course CHEM 1307 taught by Professor Li during the Spring '08 term at Texas Tech.

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1307EX2Btest - Equations: o K = oC + 273.15 o F = 1.8 (oC)...

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