Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

CHM 201-003 Sp. 17 Exam 1 Answer the questions in order in the blue book. Where

ID: 570343 • Letter: C

Question

CHM 201-003 Sp. 17 Exam 1 Answer the questions in order in the blue book. Where appropriate, you must show your calculations and give your answer to the correct number of significant digits for full credit. 1. Convert: dm (a) 0.305 kg to mg (b) 214 mm to m (c) 0.0215 L to ml (d) 1.05 L to 2. (a) What is the density of chromium metal if a piece weighing 15.0 g has a volume of 1.80 ml: and, (b) What is the volume of a piece of aluminum metal weighing 435 g? The density of Al is 2.74 g/em 3. a) Give the no. ofelectrons, protons and neutrons in each of the following nuclides: i) Na-24 ii) Sn-120 i) U-238 b) Give the number of protons and electrons in i) Cui)Pil) 4. Calculate the atomic mass of magnesium given that the metal has three naturally occurring isotopes: Mg-25 ( 10.00 % abundance, isotopic mass-24.986 amu): Mg-24 (78.99% abundance isotopic mass-23.985 amu) and Mg-23 (11.01 % abundance, isotopic mass = 22.99 amu). 5. Calculate: a) The number of calcium atoms in a 2.00 g sample of the metal b) The number of oxygen molecules in a 4.00 g sample of the gas, and; c) The mass of 3.011 x 10 molecules of N 6. Give systemic names for the following compounds: a) CriNO3)3 b) BaC12 c) MnO d) N203: e) MgSO7 1H20 7. Write the formula for each of the following compounds: a) ammonium bromide b) diphosphorus pentoxide c) potassium carbonate d) calcium hydroxide e) nickel chloride (1I) hexahydrate

Explanation / Answer

1. unit conversion

(a) 0.305 kg = 0.305 kg x 1000 g x 1000 mg = 305,000 mg

(b) 214 mm x (1/10 cm) x (1/100 m) = 0.214 m

(c) 0.0215 L x 1000 ml = 21.5 ml

(d) 1.05 L = 1.05 dm^3

--

2. (a) density of chromium = mass/volume

                                           = 15 g/1.80 ml

                                           = 8.33 g/ml

(b) Volume of Al metal = mass/density

                                     = 4.35 g/2.74 g/cm^3

                                     = 1.59 cm^3

--

3. (a) for the nuclide

i) Na-24

Electron = 11; Proton = 11; Neutron = 12

ii) Sn-120

Electron = 50; Proton = 50; Neutron = 70

iii) U-238

Electron = 92; Proton = 92; Neutron = 146

--