Honors chemistry

Fall Examination Study Questions

 

1. (Chapter 1) How many significant figures are there in the following numbers or answers?

a) 10.2 b) 0.0030 c) 3.1 x 105 d) 6.382 + 1.2 = ? e) 8.0 x 10.0 = ?

 

2. (Chapter 1) Identify the following properties as physical or chemical properties:

a) Copper is shiny and orange.

b) Potassium reacts explosively with fluorine gas to produce potassium fluoride.

c) Oxygen is a gas at room temperature.

d) Sodium oxide has a very high melting point.

e) Sodium chloride dissolves readily in water.

 

3. (Chapter 2) For each of the following compounds indicate whether it is ionic or covalent and give the correct formula. For the covalent compound(s), is the formula an empirical formula?

a) dinitrogen pentoxide b) magnesium nitrate c) silver(I)oxide d) potassium hydroxide

 

4. (Chapter 3) Rust is 52.3% Fe, 44.9% O and 2.8% H.

a) Find the empirical (simplest) formula for rust.

b) Rust is an ionic compound. What two ions are present in rust?

c) Write a balanced equation for the formation of rust from iron(III)oxide and water.

 

5. (Chapter 2) Name the following compounds:

a) PbCl2 b) Cu2SO4 c) CS2 d) HF e) NaClO3

 

6. (Chapter 3) Perform the following calculations about barium hydroxide.

a) Determine the mass percentage of each element in barium hydroxide.

b) Determine the number of moles of oxygen in 49.7 grams of barium hydroxide.

 

7. (Chap 3 & 6) For the reaction, 3 CO(g) + 7 H2(g) C3H8(g) + 3 H2O(l)

determine:

a) the mass of C3H8 produced when 3.66 moles of H2 are used up.

b) the mass of water produced from 144 milligrams of H2.

c) the number of molecules of C3H8 produced from 8.0 moles of carbon monoxide.

d) the mass of propane formed if 28.0 grams of H2 are combined with 126 grams of CO.

 

8. (Chapter 4) What mass of sodium nitrate is needed to make 157 ml of a 3.00 M sodium nitrate solution?

 

9. (Chapter 4) Silver nitrate and calcium chloride react to form a precipitate.

a) Write a balanced formula equation for this reaction.

b) What volume of a 0.120 M calcium chloride solution is needed to react with an excess of aqueous silver nitrate to form 2.89 grams of the precipitate?

c) If only 2.11 grams of precipitate are formed in (b), what is the percent yield of precipitate?

 

10. (Chap 4) List one example each of a) strong acid, b) weak acid, c) strong base, d) weak base

 

11. (Chapter 7) Give the ground state electron configuration for the following atoms or ions:

a) Ar b) Ca2+ c) Y d) F-

 

12. (Chapter 7) Which element . . . . . ?

a) has an outer electron configuration of 3s2?

b) is the least electronegative element in Period 3?

c) has the smallest atomic radius in Group 14?

 

13. (Chapter 7) Give possible values for the 4 quantum numbers for a 4f electron.

 

14. (Chapter 8) Draw Lewis structures for the following molecules:

a) CH3Cl b) N2 c) C2H4O d) H2SO4

 

15. (Chapter 22) Draw structures for each of the following compounds:

a) acetic acid b) an aromatic alcohol c) 3-ethyl, 2,3-dimethylhexane d) an isomer of (c)

 

16. (Chapter 22) Draw the polyester made from

 

 

 

 


17. (Chapter 8 & 9) For each of the following molecules, indicate the geometry and indicate whether the molecule is polar or nonpolar. (b & c have extended octets)

a) CH2Cl2 b) SCl4 c) XeF4 d) NH3

 

18. (Chapter 2 & 4) Decide whether a precipitate will form when the following solutions are mixed. If a precipitate forms, give the name and formula of the precipitate.

a) sodium sulfate and barium chloride

b) ammonium sulfide and strontium nitrate

c) lithium carbonate and cobalt(III) chloride

 

19. (Chapter 4) For each of the net reactions in Question 18, write both a balanced formula equation and a net ionic equation for the reaction.

 

20. (Chapter 3) A sample of aluminum is heated in air and completely converted to aluminum oxide. Use the data below to calculate the empirical formula of aluminum oxide.

Mass of crucible = 29.00 g

Mass of crucible + aluminum (before heating) = 30.62 g

Mass of crucible + oxide (after heating) = 32.06 g

 

 

 


Honors Chemistry

Answers to Fall Examination Study Questions

 

1. a) 3 b) 2 c) 2 d) 2 (answer is 7.6) e) 2

2. a) physical b) chemical c) physical d) physical e) physical

3. a) covalent, empirical, N2O5 b) ionic, Mg(NO3)2 c) ionic, Ag2O d) ionic, KOH

4. a) In 100 g of the compound:

52.3 g Fe x = 0.937 mol Fe 0.937/0.937 = 1 FeO3H3 =

44.9 g O x = 2.81 mol O 2.81/0.937 = 3 Fe(OH)3

2.8 g H x = 2.8 moles H 2.8/0.937 = 3

b) Fe3+ and OH- c) Fe2O3(s) + 3 H2O(l) 2 Fe(OH)3(s)

5. a) lead(II) chloride b) copper(I) sulfate c) carbon disulfide d) hydrofluoric acid

e) sodium chlorate

6. a) Ba(OH)2 = 1 mole Ba = 137.34 g Ba 137.34/171.36 = 80.1% Ba

2 mole O = 32.00 g O 32.0/171.36 = 18.7% O

2 mole H = 2.016 g H 2.016/171.36 = 1.18% H

b) 49.7 g Ba(OH)2 x = 0.580 moles O

7. a) 3.66 mol H2 = 23.0 g C3H8

b) 144 mg H2 = 0.551 g H2O

c) 8.0 moles CO = 16.0 x 1023 = 1.6 x 1024 molecules

d) limiting reactant problem:

28.0 g H2 x = 88.0 g C3H8

126 g CO x = 66.0 g C3H8, 66.0 g C3H8

 

8. 157 ml x = 40.0 g NaNO3

9. a) 2 AgNO3(aq) + CaCl2(aq) 2 AgCl(s) + Ca(NO3)2(aq)

b) 2.89 g AgCl x = 84.0 mL

c) % yield = experimental yield/theoretic yield x 100% = 2.11/2.89 x 100% = 73.0%

10. a) HCl, HNO3 b) HNO2, HF, CH3COOH c) NaOH, LiOH d) NH3, NO2-

11. a) 1s22s22p63s23p6 b) 1s22s22p63s23p6 c) 1s22s22p63s23p64s23d104p65s24d1 d) 1s22s22p6

12. a) Mg (period 3, Group 2) b) Na c) C

13. n = 4, l = 3, ml = 3, 2, 1, 0, -1, -2, or -3, ms = + or -

14.

a) b) c) d)

 

 

 


15. O CH3 CH3 CH3

|| | | |

a) CH3-C-OH c) CH3-CH-C-CH2-CH2-CH3 d) CH3-C-CH2-C-CH2-CH3

| | | |

b) -O-H CH3 CH2-CH3 CH3 CH3

 

16. O O O O O O

|| || || || || ||

-O-CH2-O-C-CH-C-O-CH2-O-C-CH-C-O-CH2-O- C-CH-C-

| | |

CH3 CH3 CH3

 

17. a) tetrahedral, polar b) see-saw, polar

c) square planar, nonpolar d) trigonal pyramid, polar

18. a) barium sulfate, BaSO4 b) no reaction c) cobalt(III) carbonate, Co2(CO3)3

19. a) Na2SO4(aq) + BaCl2(aq) 2 NaCl(aq) + BaSO4(s); Ba2+(aq) + SO42-(aq) BaSO4(s)

c) 3 Li2CO3(aq) + 2 CoCl3(aq) 6 LiCl(aq) + Co2(CO3)3(s);

2 Co3+(aq) + 3 CO32-(aq) Co2(CO3)3(s)

20. Mass of Al = 30.62 g - 29.00 g = 1.62 g Al; Mass of O = 32.06 g - 30.62 g = 1.44 g O

1.62 g Al x = 0.0600 mol Al 0.0600/0.0600 = 1 x 2 = 2

1.44 g O x = 0.0900 mole O 0.0900/0.0600 = 1.5 x 2 = 3 Al2O3