Chemistry review for Fundamentals Engineering Exam

1. Chemistry review for Fundamentals Engineering Exam

2. General Information:concentrations • Moles solute/Liter solution = Molarity • gram/106grams = parts per million • g/109g = parts per billion • g/1012g = parts per trillion • grams component/grams total x 100% = weight percent • Moles solute/kilograms solvent = molality

3. General Information:elements • atomic mass unit (a.m.u.) = the unit of mass that is used to describe the masses of atoms = equivalent to 1/(6.022x1023) of a gram. • mass number = the number of protons and neutrons in a nucleus • Atomic number = the number of protons in a nucleus • Isotopes have same atomic number but different mass numbers.

6. Metals (in red) comprise most of the elements, and are characterized by their tendency to lose electrons when involved in chemical reactions. Metals make good electrical and thermal conductors because of this property. Non-metals (in blue) have a tendency to gain electrons when in combination with metals, and are thus oxidizing agents. (The rightmost column of the Periodic Table is shown in blue, but these elements are, in fact, inert) Metalloids, semi-metals, or semiconductors can swing both ways and are found in the region of the Table between the metals and non-metals.

14. Isotope problem X (49.9472) + (1-x)(50.9440 = 50.9415 X= 0.0025, 1-x = 0.9975

15. Isotope problem

16. Periodic trends 1. Choose the larger atom in each pair: Mg, Ca; Be, F; Cl, Br; S, Cl. 2. Choose the element with the greater ionization energy in each pair: Mg, Ca; Be, F; Cl, Br; S, Cl. 3. Choose the element with the greater electron affinity in each pair: Be, F; Cl, Br; S, Cl. 4. Choose the more electronegative element in each pair: Mg, Ca; Be, F; Cl, Br; S, Cl.

17. Periodic trends Choose the larger atom in each pair: Mg, Ca; Be, F; Cl, Br; S, Cl.

18. Periodic trends Choose the element with the greater ionization energy in each pair: Mg, Ca Be, F Cl, Br S, Cl.

19. Periodic trends 3. Choose the element with the greater electron affinity in each pair: Be, F Cl, Br S, Cl.

20. Periodic trends Choose the more electronegative element in each pair: Mg, Ca Be, F Cl, Br S, Cl.

21. Periodic trends Choose the more metallic element in each pair: Mg, Ca Be, Ra Ga, Ir Li, Be.

22. Chemical Reactions

23. Gas Laws Avogadro's hypothesis, which states: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

24. Gas Laws Ideal Gas Law, shown below, relates the volume, temperature, and pressure of a gas, considering the amount of gas present. PV = nRT P=pressure in atm T=temperature in Kelvins R is the molar gas constant, where R=0.082058 L atm mol-1 K-1. V = Volume in Liters n = moles Note:

25. Gas Laws Combined Gas Law: quantitatively determine how changing the pressure, temperature, volume, and number of moles of substance affects the system. Example Problem: A 25.0 mL sample of gas is enclosed in a flask at 22°C. If the flask was placed in an ice bath at 0°C, what would the new gas volume be if the pressure is held constant? Answer:   Because the pressure and the number of moles are held constant, we do not   need to represent them in the equation because their values will cancel. So the   combined gas law equation becomes:

26. Gas Laws Combined Gas Law: quantitatively determine how changing the pressure, temperature, volume, and number of moles of substance affects the system. Also useful for partial pressures: partial pressure of each the gas in the mixture is equal to the total pressure multiplied by the mole fraction.

27. Non-Ideal Gas Law P = pressure V = volume N = number moles R = gas constant Where the van der Waals constants are: a accounts for molecular attraction b accounts for volume of molecules

28. General information on Acids/Bases • HA(aq) H+(aq) + A-(aq) • pH = -log[H+] • Ka = [H+][A-]/[HA] • pH =pKa + log([conjugate base]/[acid])

29. Determine the pH of 0.30 M acetic acid (HC2H3O2) with the Ka of 1.8x10-5.

30. 0.040 L x 0.5 Moles/Liter x 2 moles H+/H2SO4 = 0.04 moles NaOH 0.04 moles/0.06 Liters = 0.67 M NaOH

31. 0.040 L x 0.5 Moles/Liter x 2 moles H+/H2SO4 = 0.04 moles NaOH 0.04 moles/0.06 Liters = 0.67 M NaOH

32. Oxidation numbers 1) For atoms in their elemental form, the oxidation number is 0 2) For ions, the oxidation number is equal to their charge 3) For single hydrogen, the number is usually +1 but in some cases it is -1 4) For oxygen, the number is usually -2 (except in H2O2 where it is -1) 5) The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.

33. Oxidation numbers 1) For atoms in their elemental form, the oxidation number is 0 2) For ions, the oxidation number is equal to their charge 3) For single hydrogen, the number is usually +1 but in some cases it is -1 4) For oxygen, the number is usually -2 (except in H2O2 where it is -1) 5) The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge. C CO CO2 CO32-

34. Oxidation numbers 1) For atoms in their elemental form, the oxidation number is 0 2) For ions, the oxidation number is equal to their charge 3) For single hydrogen, the number is usually +1 but in some cases it is -1 4) For oxygen, the number is usually -2 (except in H2O2 where it is -1) 5) The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.

35. Oxidation numbers 1) For atoms in their elemental form, the oxidation number is 0 2) For ions, the oxidation number is equal to their charge 3) For single hydrogen, the number is usually +1 but in some cases it is -1 4) For oxygen, the number is usually -2 (except in H2O2 where it is -1) 5) The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.

36. Metal oxidation states Red means commonly found

37. Balancing Redox equations 1) Divide the equation into an oxidation half-reaction and a reduction half-reaction 2) Balance these Balance the elements other than H and O Balance the O by adding H2O Balance the H by adding H+ Balance the charge by adding e- 3) Multiply each half-reaction by an integer such that the number of e- lost in one equals the number gained in the other 4) Combine the half-reactions and cancel 5) If pH > 7, **Add OH- to each side until all H+ is gone and then cancel again**

38. Balancing Redox equations 1) Divide the equation into an oxidation half-reaction and a reduction half-reaction 2) Balance these Balance the elements other than H and O Balance the O by adding H2O Balance the H by adding H+ Balance the charge by adding e- 3) Multiply each half-reaction by an integer such that the number of e- lost in one equals the number gained in the other 4) Combine the half-reactions and cancel 5) If pH > 7, **Add OH- to each side until all H+ is gone and then cancel again**

40. Half life calculations dA Rate = - = k dt dA = -k dt A At t t [A]t dA ln = -kt = -k dt = -k dt A [A]0 A0 0 0 [A]t e-kt ½ = @ half life, ln ½ = -kt & Then, & k = 0.693/t = 1/2 [A]0 -ln(8.7mg/25 mg] [A]t (30 years) e-0.693t/t ln([A]t/[A]0] = -0.693t/t t = = 0.693 [A]0 t = 45 years

42. PV = nRT 697 mmHg V = 1.9 mol R 294K 795 mmHg V = 1.9 + n2 moles R 299K R = 62.363 x 10-3 m3mmHgK-1mol-1 Not necessary to know R! V1 = V2 Solve for n2 n2 = n1T1P2/T2P1 = (1.9 mol 294 K 795 mm)/(299K 697 mm) = 2.13 moles

44. n = 5 moles throughout Temp = constant. P1V1 = P2V2 P1 = 1 atm, V1 = 10 L, V2 = 15 L P2 = P1V1/V2= 1 atm x 10 L/15 L = 0.67 atm.

46. M = moles/Liter Molecular weight of CH3OH = 12.01 + 4(1.00794) + 15.99994 = approx 32 # moles = 11.7g/32 = 0.365 moles M = 0.365 moles/0.230 Liters = 1.59 M

48. The number of oxygen atoms in 100 grams of calcium carbonate is? • 300 • 6.022 x 1023 • 16.066 x 1023 • 30,000,000

49. The number of oxygen atoms in 100 grams of calcium carbonate is? • 300 • 6.022 x 1023 • 16.066 x 1023 • 30,000,000

50. Ten moles of Ca(NO3)2, in grams, is: • 1020 • 1100 • 1640 • 1800