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Acids and bases

Acids and bases. Pages 21, 22, 23. 76. Complete the chart. 77. Write the steps to name an acid. Hydrogen + a polyatomic? Polyatomic ends in –ate the acid ends in – ic Polyatomic ends in – ic the acid ends in – ous Hydrogen + a single element? Hydro______ ic acid.

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Acids and bases

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  1. Acids and bases Pages 21, 22, 23

  2. 76. Complete the chart

  3. 77. Write the steps to name an acid • Hydrogen + a polyatomic? • Polyatomic ends in –ate the acid ends in –ic • Polyatomic ends in –ic the acid ends in –ous • Hydrogen + a single element? • Hydro______ ic acid

  4. 78. Name the following acids a) HCl – Hydrochloric acid b) HF – Hydrofluroic acid c) HNO2- Nitrous acid d) H2SO4 – Sulfuric acid

  5. 79. List the steps for formula an acid • If the acids ends in an “ic” then the polyatomic ends in the –ate form • If the acids ends in “ous” then the polyatomic is in the – ite form If the acid starts in Hydro- then the formula is Hydrogen followed by the element ending in –ide Hydrochloric acid= HCl

  6. 80. Write the formula for the following acids: a) nitric acid - HNO3 b) acetic acid – HC2H3O2 c) hydrobromic acid - HBr d) sulfurous acid – H2SO3

  7. 81. Write the steps naming a base. • Following ionic naming rules Name the cation regularly, the polyatomic uses its regular name.

  8. 82. Name the following bases: a)KOH - Potassium hydroxide b) Ba(OH)2- Barium hydroxide c) LiOH- Lithium hydroxide d) NH3 – ammonia

  9. 83. List the steps for writing the formula for a base. • Use ionic rules Write the symbols. Identify the charges- criss cross to make subscripts.

  10. 84. Write the formula for the following bases: • a) Sodium hydroxide NaOH • b) Beryllium hydroxide Be(OH)2 • c) Calcium hydroxide Ca(OH)2 • d) Cesium hydroxide CsOH

  11. 85. What is the difference between the Arrhenius and Bronsted-Lowry definitions of an Acid? • Arrhenius acid- contains an H+ • Bronsted- Lowry acid- Donates an H+

  12. 86. What is the difference between the Arrhenius and Bronsted-Lowry definitions of a Base? • Arrhenius base- contains an OH- • Bronsted- Lowry base- Accepts an H+

  13. 87. Identify the Bronsted-Lowry Acid and Base on the left side of the equation and identify the conjugateacid-base on the right side of the equation. • HNO3 + H2O  H3O+ + NO3- ACIDBASE ACID BASE • N2H4 + H2O  N2H3- + H3O+ ACID BASE BASE ACID • N2H4 + HCl N2H5+ + Cl- BASE ACID ACID BASE

  14. 88. Determine the products for the following neutralization reactions (double replacement) • Neutralization reactions: ACID+ BASE  SALT + WATER KOH + H2CO3 KOH + H2CO3 H2O + K2CO3 Don’t forget to BALANCE! 2KOH + H2CO3 2H2O + K2CO3 HBr + Al(OH)3 HBr + Al(OH)3 H2O + AlBr3 3HBr + Al(OH)3 3H2O + AlBr3

  15. 89. Draw the pH scale indicating concentrations of H+ and OH- ions.

  16. 90. Complete the table for the missing information:

  17. 91. What is a titration and what is it used for? • Titration- method for determining concentration of a solution by reacting a known volume of solution with a solution of known concentration • Equivalence point- equal amounts of OH- & H+ ions

  18. Nuclear Pages 24

  19. 94. Chemical and Nuclear Reaction

  20. 93. 3 Types of Nuclear Radiation +2 -1 0 Helium nucleus electron EMR 42a or42He0-1b or0-1e g 4 amu1/1840 amu 0 low medium high Paper, cloth, skin, etc. Aluminum foil Lead

  21. 94. Write a balanced nuclear equation for the followinga. Au-191 undergoes beta decay: • Beta decay = electron • 191 is gold’s Atomic mass, which goes on top. • Look up Gold (Au) on your periodic table to find the atomic number which goes on the bottom. Gold’s atomic number is 79. • 19179Au  0-1e +_____ • 19179Au  0-1e + 19180Hg

  22. 94. Write a balanced nuclear equation for the following b. Rb-90 undergoes alpha decay: • alpha decay = helium particle • 90 is Rubidium’s Atomic mass, which goes on top. • Look up Rubidium (RB) on your periodic table to find the atomic number which goes on the bottom. Rubidium’s atomic number is 37. • 9037Rb  42He +_____ • 9037Rb  42He + 8635Br

  23. 95. Find the missing particle A. 42He + 147N  _____ + 11H 42He + 147N 176C+ 11H B. 10244Ru + 42He10n +_____ 10244Ru + 42He10n + 10546Pd

  24. Thermochemistry Page 25

  25. 96. What is the difference between potential and kinetic energy? • Potential energy- stored energy due to position • Kinetic energy- energy of motion *remember that Temperature is a measure of the average kinetic energy

  26. 97. Explain the difference between heat and temperature. • Heat- (Q) = the process of flowing from warmer to colder temperature. • Temperature- measure of the average kinetic energy (KE) in a sample

  27. 98. What is specific heat? What is the SI unit? • Specific heat (c) is the amount of energy required to raise the temperature of a 1 gram sample by 1 degree • High specific heat means that the substance warms and cools slowly. It resists changes in temperature. • Low specific heat means that the substance warms and cools quickly. • SI Unit = J/(g·°C)

  28. 99. What are the 3 ways heat/energy is transferred? Give an example for each type? a. Conduction- heat is transferred by touch. Ex: heating a pan on the stove b. Convection- heat is transferred through liquids or gases. Ex: Cooking in an oven c. Radiation- heat from the sun

  29. 100. Indicate whether the energy change is endothermic or exothermic for the following: • a. -200 kJ exothermic • b. 32 kJ endothermic • c. 653.8368 kJ endothermic Exothermic reactions release energy so they lose energy (negative sign) Endothermic reactions absorb heat so they have energy added (positive sign)

  30. 101. The specific heat of iron metal is .450 J/g · C. How many Joules of heat are necessary to raise the temperature of a 1.05 g block of Iron from 25 C to 88.5 CIs this process endothermic or exothermic? FORMULA: Q=mc∆T Q= heat(J); m= mass(g); c=specific heat J/(g·°C); ∆T = change in temperature (Final– initial) Q= (1.05)(.450)(63.5) Q= 30.0 J The reaction is endothermic because it absorbs heat. ∆T = 88.5-25 = 63.5

  31. 102. Calculate the amount of heat produced when 52.4 g of methane, CH4, burns in an excess of air,according to the following equation. Balance the equation: 1CH4(g) + 2O2(g)1CO2(g) +2H2O(l) Hrxn= -890.2 kJ/mole 52.4g 1 mol =3.27 moles 16.043 g 3.27 mol -890.2kJ = -2910 kJ 1 mole

  32. Gas Laws Page 26, 27, 28

  33. 103. What are the 4 variables (including units) that affect the behavior of a gas? 1. Pressure (P)- atm, torr, kpa, mmHg, psi 2. Volume- (V) Liters 3. Temperature (T)- Kelvin 4. Amount- (n) moles R = gas constant!

  34. 104. List the postulates of the kinetic molecular theory. 1. Gases consist of molecules whose separation is much larger than the size of the molecules themselves. 2. Particles in a gas move in straight line paths and random directions. 3. Particles in a gas collide frequently with the sides of the container and less frequently with each other. All collisions are elastic (no energy is gained or lost as a result of the collisions). 4. Particles in a gas do not attract or repel one another. They do not sense any intermolecular forces.

  35. 105. What conditions are represented by the abbreviation STP? Where can you easily find these valueswhen testing? • STP = 0 0C and 1 atm • Temperature in Kelvin = 273K STP can be found on the STAAR chemistry reference chart under the constants and conversions section.

  36. 106. List the 6 gas laws and their formulas.

  37. 107. What volume will 1.02 mole of carbon monoxide gas occupy at STP? • STP= Standard temperature and pressure P= 1 atm, T= 0 oC or 273K PV=nRT (1atm)(V)= (1.02moles)(.0821)(273) V= 22.8L

  38. 108. How many grams of helium gas are in a 1.5 L balloon at STP? • STP= Standard temperature and pressure P= 1 atm, T= 0 oC or 273K PV=nRT (1atm)(1.5L)=(n)(.0821)(273) 1.5= 22.4n n= .067 moles .067 moles 4.003g = .27 grams 1 mole Molar mass of helium

  39. 109. What temperature is required for 0.0470 moles of gas to fill a balloon to 1.20 liters under 0.988 atm pressure • PV=nRT (.988atm)(1.20L)=(.0470)(.0821)(T) 1.1856= .00385T 308K=T

  40. 110. Calculate the volume that a 3.58 g sample of neon gas will occupy at 287 K and a pressure of 0.900 atm Convert grams to moles to plug into the ideal gas law equation! 3.58g 1 mol =.177 mol 20.180g PV=nRT (.900atm)(V)=(.177mol)(.0821)(287) V= 4.63L

  41. 111. In one city, a balloon with a volume of 6.0 L is filled with air at 101 kPa pressure. The balloon is thentaken to a second city at a much higher altitude. At this second city, atmospheric is only 91 kPa. If thetemperature is the same in both places, what will be the new volume of the balloon? • Comparing Volume and Pressure – Boyles’ law V1P1=V2P2 (6L)(101kPa)=(V2)(91kPa) 606=(V2)(91kPa) 6.7L=(V2)

  42. 112. A certain mass of gas in a 2.25 L container has a pressure of 164 kPa. What will the new pressure beif the volume of the container is reduced to 1.50 L and the temperature stays constant? • Comparing Volume and pressure- Boyles’ law V1P1=V2P2 (2.25L)(164kPa)=(1.50L)(P2) 369= (1.50L)(P2) 246 kPa=(P2)

  43. 113. A 500 K, a 10.5 L sample of gas has a pressure of 0.948 atm. The gas is allowed to expand into a 25.0 L container at 618 K. What is the new pressure? • Comparing Temperature, volume and Pressure – Combined gas law P1V1 = P2V2 T1 T2 • (10.5)(.948) = (25.0)(P2) 500 618 • .019908= (25.0)(P2) 618 • 12.3= (25.0)(P2) • P2 = .49 atm

  44. 114. A 7.36 L sample of gas has a temperature of 50 oC. In order to keep the pressure constant, what volume must the gas occupy at –100 oC? • Comparing volume and temperature- Charles (7.36L) = (V2) (323K) (173K) (V2)= 3.94L

  45. 115. A steel container has the following mixture of gases: 1.2 atmospheres of Xenon, 0.75 atmospheres of Chlorine, and 0.41 atmospheres of Neon. What is the total pressure of gases in the container? • Dalton’s law of partial pressure *Be sure that all of the pressure values have the same units. PTOT = P1+ P2+P3 PTOT = 1.2atm +.75atm +.41atm PTOT = 2.36 atm

  46. 116. A scuba tank contains a mixture of 0.59 atm of Oxygen, 0.65 atm of Nitrogen, and an unknownamount of carbon dioxide. If the total pressure in the tank is 199 kPa, what is the pressure of thecarbon dioxide in atm? • Dalton’s law of partial pressure *Be sure that all of the pressure values have the same units. Convert kPA to atm 101kPa=1atm: 199 kPa 1 atm = 1.97atm 101kPa PTOT = P1+ P2+P3 1.97= .59+.65 +P3 .73atm =P3

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