Chemistry 123 Fall ’14 Exam I (CH I‚ II‚ III) Instructions: Read each problem carefully before you begin. Be certain that you answers are clear and legible: Clearly Circle One Answer Only. Make sure to review you answers before you turn the exam in. Please place your answers on the answer sheet. You should also circle the correct letter for back up purposes. Print your name: ___________________ Sign your name: ___________________ Answer Sheet 1 C 16 A 2 E 17 D 3 B
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Honors Chemistry Final Exam Study Guide Multiple Choice- (51 points @ 1/2 point ea.) Identify the letter of the choice that best completes the statement or answers the question. ____ 1. The pair of elements that forms a bond with the least ionic character is |a. |Na and Cl. |c. |O and Cl. | |b. |K and Cl. |d. |Mg and Cl.
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Test #3 Chapter 2: 1) What is matter? 2) What is mass? 3) Matter that has a uniform and definite composition is called a ______. Give an example. 4) A quality/condition of a substance that can be observed/measured without changing the substance is a ________ __________. Give examples. 5) What are the three states of matter? Put the correct state of matter in each box. 6) What is the difference between gas and vapor? 7) What is a physical change? Give an example. 8) A physical blend
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of water in barium chloride dihydrate. Solution: The formula mass of BaCl2(2H2O is Ba 1 x 137.3 = 137.3 amu Cl 2 x 35.5 = 71.0 amu H2O 2 x 18.0 = 36.0 amu 244.3 amu The theoretical percentage of water is found by dividing the water of crystallization mass‚ 36.0 amu‚ by the hydrate mass‚ 244.3 amu. 36.0 amu x 100 = 14.7 % 244.3 amu Problem 2 – Experimental percentage The experimental percentage of water in a hydrate is found by comparing the
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26 Tantalum-181: Electrons: 73‚ Protons: 73‚ Neutrons: 108 76. Ga-69‚ because its atomic mass is closer to the atomic mass of gallium‚ than Ga-71 is. 78. Cr-50 = 0.0435 x 49.946 amu = 2.17 Cr-52 = 0.8379 x 51.941 amu = 43.52 Cr-53 = 0.0950 x 52.941 amu = 5.03 Cr-54 = 0.0236 x 53.939 amu = 1.27 Total = 51.99 amu 80. When the neutron-to-proton ratios is too big or too small‚ it would cause an element to become radioactive 82. An alpha particle is a helium atom with a 2+ charge‚ a beta particle
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Mass1 = mass of first isotope in amu Abundance1 = abundance of first isotope as a percentage n = number of isotopes of element in calculation For example: The natural abundance for boron isotopes is 19.9% 10B (10.013 amu*) and 80.1% 11B (11.009 amu*). Calculate the atomic mass of boron: Average atomic mass = = 10.811 (note that this is the value of atomic mass given on the periodic table) *amu is the atomic mass unit (u‚ μ or amu)‚ which is defined as 1/12th the
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the average mass of a boron atom is 10.81 amu. Assuming you were able to isolate only a single boron atom‚ the chance that you would randomly get one with a mass of 10.81 amu is A. 0% B. 0.81% C. about 50% D. 10.81% E. 100% 5. Match the fundamental charge and mass with the appropriate subatomic particle: Charge = –1 Mass ≈ 0 amu proton proton electron electron neutron Charge = +1 Mass ≈ 1 amu neutron electron proton neutron proton Charge = 0 Mass ≈ 1 amu electron neutron neutron proton electron
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Chapter 3 1. 1 atom of X = atomic mass (amu) 1 mole f X=atomic mass in grams 1 mole of something = 6.022 x 1023 units of that substance 1 mole of a compound =66.022 x 1023 atoms 6.022 x 1023 amu =1 g 2. What is the mass of 6 atoms of Fe? Answer: 6 atoms of Fe x 55.85 amu÷atom of Fe x 1 g of Fe÷6.022 x 1023 amu = 3. How many atoms does it take to make 1 g of Gold (Au)? Answer: 197.0 g Au =1 mole of Au
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Exam 1A-CHEM 1150 F12 Name___________________________________ MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Which of the following elements is a metal? A) As B) Se C) Sn 1) D) C E) I 2) Determine the mass of an object that has a volume of 88.6 mL and a density of 9.77 g/mL. A) 298 g B) 568 g C) 907 g D) 1100 g E) 866 g 2) 3) A student performs an experiment to determine the density of a sugar solution. She
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produced? A) 48.8 g B) 32.5 g C) 131.2 g D) 73.2 g E) 65.1 g 2. Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu‚ 92.21% abundance)‚ X-29 (28.976 amu‚ 4.70% abundance)‚ and X-30 (29.974 amu‚ 3.09% abundance). Calculate the atomic weight of X. A) 28.09 amu B) 35.29 amu C) 28 amu D) 86.93 amu E) 25.80 amu 3. The Claus reactions‚ shown below‚ are used to generate elemental sulfur from hydrogen sulfide. 2H2S + 3O2 2SO2 + 2H2O SO2 + 2H2S 3S
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