Presentation is loading. Please wait.

Presentation is loading. Please wait.

Pages 21, 22, 23. PropertiesAcidsBases TasteSourBitter FeelSlippery ConductivityYes Reacts with metals (yes or no?) what products? Yes- forms H 2 gasNo.

Similar presentations


Presentation on theme: "Pages 21, 22, 23. PropertiesAcidsBases TasteSourBitter FeelSlippery ConductivityYes Reacts with metals (yes or no?) what products? Yes- forms H 2 gasNo."— Presentation transcript:

1 Pages 21, 22, 23

2 PropertiesAcidsBases TasteSourBitter FeelSlippery ConductivityYes Reacts with metals (yes or no?) what products? Yes- forms H 2 gasNo pH range

3  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 a) HCl – Hydrochloric acid b) HF – Hydrofluroic acid c) HNO 2 - Nitrous acid d) H 2 SO 4 – Sulfuric acid

5  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 a) nitric acid - HNO 3 b) acetic acid – HC 2 H 3 O 2 c) hydrobromic acid - HBr d) sulfurous acid – H 2 SO 3

7  Following ionic naming rules Name the cation regularly, the polyatomic uses its regular name.

8 a)KOH - Potassium hydroxide b) Ba(OH) 2 - Barium hydroxide c) LiOH- Lithium hydroxide d) NH3 – ammonia

9  Use ionic rules Write the symbols. Identify the charges- criss cross to make subscripts.

10  a) Sodium hydroxide NaOH  b) Beryllium hydroxide Be(OH) 2  c) Calcium hydroxide Ca(OH) 2  d) Cesium hydroxide CsOH

11  Arrhenius acid- contains an H+  Bronsted- Lowry acid- Donates an H+

12  Arrhenius base- contains an OH-  Bronsted- Lowry base- Accepts an H+

13  HNO 3 + H 2 O  H 3 O + + NO 3 - ACID BASE ACIDBASE  N 2 H 4 + H 2 O  N 2 H H 3 O + ACID BASE BASEACID  N 2 H 4 + HCl  N 2 H Cl - BASE ACID ACID BASE

14  Neutralization reactions: ACID+ BASE  SALT + WATER KOH + H 2 CO 3  KOH + H 2 CO 3  H 2 O + K 2 CO 3 Don’t forget to BALANCE! 2KOH + H 2 CO 3  2H 2 O + K 2 CO 3 HBr + Al(OH) 3  HBr + Al(OH) 3  H 2 O + AlBr 3 3HBr + Al(OH) 3  3H 2 O + AlBr 3

15

16 [H+]pH[OH-]pOHAcid, base, neutral? 1.2.3x10 -5 M x M9.36Acid 2.5.0x10 -8 M7.32.0x10 -7 M6.7Base 3.1.2x M x10 -5 M3.1Base

17  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 Pages 24

19 94. Chemical and Nuclear Reaction

20 Radiation Type AlphaBetaGamma Charge Description Symbol Mass Penetrating Power Shielding Needed 4 2  or 4 2 He 0 -1  or 0 -1 e  low medium high Paper, cloth, skin, etc. Aluminum foil Lead Helium nucleus electron EMR 4 amu 1/1840 amu 0

21  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.  Au  0 -1 e +_____  Au  0 -1 e Hg

22  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.  Rb  4 2 He +_____  Rb  4 2 He Br

23 A. 4 2 He N  _____ H 4 2 He N  17 6 C H B Ru He  1 0 n +_____ Ru He  1 0 n Pd

24 Page 25

25  Potential energy- stored energy due to position  Kinetic energy- energy of motion *remember that Temperature is a measure of the average kinetic energy

26  Heat- (Q) = the process of flowing from warmer to colder temperature.  Temperature- measure of the average kinetic energy (KE) in a sample

27  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 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  a kJ exothermic  b. 32 kJ endothermic  c kJ endothermic Exothermic reactions release energy so they lose energy (negative sign) Endothermic reactions absorb heat so they have energy added (positive sign)

30 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 = = 63.5

31 Balance the equation: 1CH4(g) + 2O2(g)  1CO2(g) +2H2O(l) Hrxn = kJ/mole 52.4g 1 mol=3.27 moles g 3.27 mol kJ = kJ 1 mole

32 Page 26, 27, 28

33 1. Pressure (P)- atm, torr, kpa, mmHg, psi 2. Volume- (V) Liters 3. Temperature (T)- Kelvin 4. Amount- (n) moles R = gas constant!

34 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  STP = 0 0 C and 1 atm  Temperature in Kelvin = 273K STP can be found on the STAAR chemistry reference chart under the constants and conversions section.

36 LAW INDEP/DEP VARIABLES CONTROL VARIABLES MATH RELATIONSHIP FORMULA BOYLE’S V, PT, n ↑V, ↓ P inverse, indirectV 1 P 1 =V 2 P 2 CHARLES’S T, VP, n ↑T ↑V direct V 1 = V 2 T 1 T 2 GAY LUSSAC’S P, TV, n ↑T ↑P direct P 1 = P 2 T 1 T 2 AVOGADRO’S V, nP, T ↑n ↑V direct V 1 = V 2 n 1 n 2 COMBINED V, P, T,NA P 1 V 1 = P 2 V 2 T 1 T 2 IDEALP,V,n, R, TRPV=nRT

37  STP= Standard temperature and pressure P= 1 atm, T= 0 o C or 273K PV=nRT (1atm)(V)= (1.02moles)(.0821)(273) V= 22.8L

38  STP= Standard temperature and pressure P= 1 atm, T= 0 o C or 273K PV=nRT (1atm)(1.5L)=(n)(.0821)(273) 1.5= 22.4n n=.067 moles.067 moles4.003g=.27 grams 1 mole Molar mass of helium

39  PV=nRT (.988atm)(1.20L)=(.0470)(.0821)(T) =.00385T 308K=T

40 Convert grams to moles to plug into the ideal gas law equation! 3.58g1 mol =.177 mol g PV=nRT (.900atm)(V)=(.177mol)(.0821)(287) V= 4.63L

41  Comparing Volume and Pressure – Boyles’ law V 1 P 1 =V 2 P 2 (6L)(101kPa)=(V 2 )(91kPa) 606=(V 2 )(91kPa) 6.7L=(V 2 )

42  Comparing Volume and pressure- Boyles’ law V 1 P 1 =V 2 P 2 (2.25L)(164kPa)=(1.50L)( P 2 ) 369= (1.50L)( P 2 ) 246 kPa=( P 2 )

43  Comparing Temperature, volume and Pressure – Combined gas law P 1 V 1 = P 2 V 2 T 1 T 2  (10.5)(.948) = (25.0)(P 2 )  = (25.0)(P 2 ) 618  12.3= (25.0)(P 2 )  P 2 =.49 atm

44  Comparing volume and temperature- Charles (7.36L) = ( V 2 ) (323K) (173K) ( V 2 ) = 3.94L

45  Dalton’s law of partial pressure *Be sure that all of the pressure values have the same units. P TOT = P 1 + P 2 +P 3 P TOT = 1.2atm +.75atm +.41atm P TOT = 2.36 atm

46  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 kPa1 atm = 1.97atm 101kPa P TOT = P 1 + P 2 +P = P 3.73atm =P 3


Download ppt "Pages 21, 22, 23. PropertiesAcidsBases TasteSourBitter FeelSlippery ConductivityYes Reacts with metals (yes or no?) what products? Yes- forms H 2 gasNo."

Similar presentations


Ads by Google