CHEM 115 EXAM #1 Practice, Fall 2011

Transcription

1 Name CHEM 115 EXAM #1 Practice, Fall 011 Circle the correct answer to each question. (numbers 1-7, points each) 1. The term used to describe a substance composed of two or more elements chemically combined in a fixed ratio is: a. solution b. compound c. heterogeneous mixture d. homogeneous mixture e. aggregate. What is not the same in different isotopes of the same element? a. atomic number b. atomic mass c. number of electrons d. number of neutrons e. mass number 3. Which of the following is a compound? a. CH 4 b. FeCl c. CO d. MgO e. none of the above f. all of the above 4. Which of the following elements is a transition metal? a. sodium, Na b. argon, Ar c. cesium, Cs d. iron, Fe e. barium, Ba 5. Which of the following elements has chemical properties similar to those of oxygen, O? a. Iron, Fe b. sulfur, S c. magnesium, Mg d. fluorine, F e. potassium, K 6. Which of the following elements has chemical properties similar to those of lithium, Li? a. Iron, Fe b. oxygen, O c. magnesium, Mg d. chlorine, Cl e. potassium, K 7. Which of the following is the smallest distance? [NOTE: all values have 3 sig. figs.] a µm b x 10-5 cm c x 10-4 mm d x 10 3 nm e. 700 pm (6 points) 8. Specify the numbers of protons, neutrons, and electrons found in 56 Fe and 56 Fe 3+. #n #p #e 56 Fe Fe

2 Fill in the blank with the appropriate word, phrase, or numerical answer. (numbers 9-1, points for each blank) 9. Matter commonly is found in three physical states. List the two that would be expected to have similar densities? liquid and solid. 10. A mixture that has different regions of composition and properties is known as a heterogeneous mixture. 11. A scientist obtains the number on a calculator. If this number actually has five (5) significant figures, how should it be written? or x Round to three significant figures and express in scientific notation. 3.4 x 10-3 (15 points) 13. a. Lead (Pb) has a density of g /cm3. Calculate the mass of a block of lead that has dimensions m x m x m. I started by converting each length to cm b/c this made life easy (just my personal opinion) Volume = cm x cm x cm = 0.30 cm 3 (Density)x(Volume) = mass, mass = (11.34 g /cm3)*(0.30 cm 3 ) mass = g b. Water has a density of 1.00 g /cm3. Find the volume of water has the same mass as the lead block in part a. volume = mass / density = (3.6 9 g)/ (1.00 g /cm3) = cm 3 c. Comment on the volume of lead and water (hint: it is helpful to look at the ratio of the volume of water / volume of lead). Volume Ratio = 3.69 / 0.30 = cm3 of water / cm3 of lead Since the water is times less dense than lead, in order to have equal masses, a volume of water must be times the volume of lead.

3 3 (5 points) 14. Define a compound and make sure to explain clearly how it differs from a homogeneous mixture by giving examples of both. A compound is a chemical combination of two or more elements in a fixed ratio (ex. include NaCl, H O, CuSO 4 and many, many more). A homogenous mixture is a physical combination of two or more substances (which can be either elements or compounds) that can be made in varying ratios (ex. include clean air, sugar completely dissolved in water, HCl dissolved in water, gasoline, or Listerine mouthwash). (10 points) 15. (a) grams of a gaseous compound composed of ONLY C and H is burned in pure oxygen and all the water and carbon dioxide that forms is collected. If the masses collected are gram of CO and 0.11 gram of HO, what is the empirical formula of the compound? [the molar masses (in g /mole) are C = 1.01, H = 1.008, CO = 44.01, and HO = 18.0] Find the moles of carbon and hydrogen from the mass of CO and H O respectively. Then divide by the smaller (if there is a smaller answer). 1mole C 1mole CO 1mole CO x 44.01g CO x0.5961g CO = mole C mole H 1mole H O 1mole H O x x0.11g H O = mole H 18.0 g H O The ratio is 1 : 1, so the empirical formulas is CH (b) The molar mass of this compound is g/mole, what is the molecular formula? Show how you arrive at your answer. The empirical unit has a mass of ~ 13.0 and the whole molecule has a mass of / therefore, it takes six empirical units to build the molecule. 6 (CH) units can be written as C 6 H 6 (8 points) 16. When you purchase your new computer you notice that the highest safe operating temperature is listed as 30 K. Express this temperature in both ºC and ºF. C = K = (rounds to 47 C) F = 1.8( C) +3 = 116 F (note, I used in this second calculation and then rounded) 3

4 4 (10 points) 17. (a) Define and give an examples for the following: Molecular formula: indicates the exact number of atoms of each type that make up a given molecule. (ex. ethanol is made from C, 6 H, and 1 O... the molecular formula is C H 6 O) Empirical formula: The empirical formula is the smallest whole number ratio of atoms that can represent the composition of a compound. Sometimes this will be the same as the molecular formula (see the example above) and sometimes it will not (see the example below). (b) Describe the relationship between the mass of an empirical unit and a molecule (a specific example will enhance your answer). The molecule will be made up of some whole number multiple of the empirical unit. Well, I suppose the example from number 15 is as good as any. C 6 H 6 as a molecule can be represented as CH in terms of empirical formula. 18. Write the names or formulas as needed. (4 points, 1.5 point each) a. lithium perchlorate LiClO 4 b. phosphoric acid H 3 PO 4 c. calcium carbide this is an unusual case, we can discuss it CaC d. copper (II) fluoride CuF e. iron (III) sulfate Fe (SO 4 ) 3 f. diphosphorus pentoxide P O 5 g. ammonia NH 3 h. water H O h. CoCl 6 H O cobalt (II) chloride hexahydrate i. Ni 3 (PO 4 ) nickel (II) phosphate j. CO carbon monoxide k. Fe O 3 iron (III) oxide l. ZnSO 4 zinc sulfate (note: zinc is +) m. SF 6 sulfur hexafluoride n. CH 3 CH CH CH 3 butane o. H O hydrogen peroxide 4

5 5 BONUS (for up to 5 points) Sulfur forms three compounds with fluorine: SF, SF 4, SF 6. Find the mass ratios F to S in each of these compounds and discuss how these numbers seem to relate to one another. [F has a mass of u and S is 3.06 u]. I will just round off the mass of S to 3... we ll still get the message For SF the ratio F : S = 38:3 = 1.19 For SF 4 the ratio F : S = 76:3 =.38 For SF 6 the ratio F : S = 114:3 = 3.56 Since the S has, 4, and 6 F bound to it in SF, SF 4, and SF 6 respectively, the relative mass of F in the compounds increases in a predictable way. The mass of F to S doubles going from SF to SF 4 and triples going from SF to SF 6 (of course it only increases by 50% going from SF 4 to SF 6 ). 5