CHEM 115 EXAM #1 Practice Fall 2007

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1 Name CHEM 115 EXAM #1 Practice Fall 007 Circle the correct answer or fill in the blanks. (numbers 1 -,.5 points each) 1. The term used to describe a substance composed of two or more elements combined in a fixed ratio is: a. solution b. homogeneous mixture c. heterogeneous mixture d. compound e. aggregate. What do all isotopes of an element have in common? a. atomic number b. atomic mass c. number of nucleons d. number of neutrons e. mass number Fill in the blank with the appropriate word, phrase, or numerical answer. (numbers 3-7, points for each blank) 3. Matter commonly is found in three physical states. List the two that would be expected to have similar densities? SOLID and LIQUID.. A mixture that has different regions of composition and properties is known as a HETEROGENEOUS mixture. 5. A scientist obtains the number on a calculator. If this number actually has four () significant figures, how should it be written? 16.7 or Round to four significant figures and express in scientific notation What is the empirical formula of C 6 H 6? CH (5 points, 1 point per blank) 8. Use your periodic table to answer the following (use chemical symbols not names) a. the element expected to be chemically most similar to oxygen sulfur, S b. an element that has properties between metals & nonmetals several possible (ex. Si) c. an element that is a transition metal many possible (ex. Fe, Mn, Ru) d. a halogen several possible (ex. Cl) e. an alkaline earth metal several possible (ex. Ba)

2 (8 points) 9. Write the formula for water and calculate the percent by mass of the elements found in water. FORMULA H O PERCENT by MASS work shown here: *1.008gH mass % H = 100 * = 100 * = 11.19% H by mass 18.0gH masso 16.00gO mass % O = 100* = 100* = 88.79% O by mass 18.0gH CHECK: sum = ~ 100 (within rounding error) (6 points) 10. Specify the numbers of protons, neutrons, and electrons found in 56 Fe and 56 Fe 3+. #n #p #e Use periodic table... Fe is #6 56 Fe Fe (8 points) 11. Calculate the number of moles of 3.01 x 10 5 atoms of aluminum (Al) 5 molesal = x3.01 atoms = molesal atoms AND the total mass of those atoms in kilograms. 1kg 6.98gAl massal = x x9.98 moleal = kg Al 1,000g (5 points) 1. You are given a mixture of water, sand, and sugar. Discuss how you might separate this heterogeneous mixture into individual substances. First the sand may be separated by filtration (since sand is a solid). This sand should be rinsed with additional water to remove the solution (homogenous mixture) of sugar and water off of the sand in the filter paper. The sand may be dried to remove excess water. To separate the sugar and water we can gently heat this solution (homogenous mixture) to remove the water. We should take care not to overheat the mixture as the sugar could caramelize (making a sticky mess). The dried sugar will remain in the heated vessel. If we wish to reclaim the water that is being removed during this heating process, we can condense it on a cold surface (for example, inside a cold condenser) and collect the liquid.

3 3 13. Write the names or formulas as needed. (1 points, 1.5 point each) a. sodium hypochlorite NaClO b. phosphoric acid H 3 PO c. copper (II) bromide CuBr d. iron (III) sulfate Fe (SO ) 3 e. diphosphorus pentoxide P O 5 f. hydrochloric acid HCl g. ammonia NH 3 h. FeI 3 6 H O iron (III) iodide hexahydrate or ferric iodide hexahydrate i. Ni 3 (PO ) nickel (II) phosphate or nickelous phosphate j. CO carbon monoxide k. Fe O 3 iron (III) oxide or ferric oxide l. PCl 5 phophorus pentachloride m. H O water or dihydrogen monoxide n. HClO percholoric acid (1 points, 3 points each) 1. Balance the following reactions: (a) 3 NaOH (aq) + H 3 PO (aq) Na 3 PO (aq) + 3 H O (l) (b) Pb(NO 3 ) (aq) + KI (aq) KNO 3(aq) + PbI (s) (c) C 3 H 8 (g) + 5 O (g) 3 CO (g) + H O (g)

4 (d) Fe (s) + 3 Cl (aq) FeCl 3 (s) (18 points) 15. CH (g) + O (g) CO (g) + H O (l) a. balance the above reaction properly and write in the following space: CH (g) + O (g) CO (g) + H O (l) b. If 30.0 g of CH reacts with 18.0 g of O, how much H O (in grams) can be formed? Find moles of both reactants, determine which is the limiting reagent, compute mass of water. 1mole = x30.0gch = g 1mole moleo = x18.0go =. 000moleO 3.00g Since, we need mole of O for every 1 mole of CH, we can see that the O is present in excess because.000 / =.1. Thus the ratio of O / CH in our mixture is >. CH is LIMITING in this reaction with these starting quantities. 18.0gH O moleh O 87 = x x1. 0 1moleH 1 = g H O c. If the actual yield is g of H O, then what is the percentage yield? actual yield 66.98gH % yield = 100x = 100x = % theoretical yield gh This is nearly 100% and perhaps the difference is due to poor trapping of the H O, an error in the mass of the starting CH, an error in the mass of the trapped H O, incomplete reaction, or some combination of these.

5 5 BONUS!!!!!! (for up to 5 points complete the following). Avogadro s number is huge. To try to understand how big this number really is consider the following: A basketball has a diameter of 9. inches. The Earth has a surface area of 196,935,000 sq miles. How many basketballs are required to cover the Earth with a single layer of basketballs? (note: if your calculator does not have a key for π, try using / 7 ) Find the cross-sectional area of a B-ball (I chose to do this in sq. miles) Area = π = π r 1mile 1,760yd 1yd 9.in x x 36in = 1.7 Divide the surface area of the Earth by the cross-sectional area of the B-ball (note: I ignored the small gaps that will exist between the balls see the section on crystal packing in your text). 196,935,000mi 16 # B balls = = mi 3 8 mi What amount in mole(s) does the number of basketballs represent? 1 8 mole 16 mole = x1.1 = mole WOW! What mass of aluminum contains a number of aluminum atoms equal to the number of basketballs found above? massal = 6.98gAl 8 x1.89 moleal = gal... or 0.51 µg WOW! That means that the number of atoms of Al present in just ~ ½ micro-gram of Al is equal to the number of basketballs you would need to blanket the surface of the Earth! That s a very small sliver of Al....and a very large number of basket-balls.

CHEM 115 EXAM #2 Practice Fall 2013

Name CHEM 115 EXAM #2 Practice Fall 2013 Circle the most correct answer for the following. (3 points each) 1. Which of the following electron configurations is not allowed by the Pauli exclusion principle?