Entry Task: June 11 th Review Ch, 15, 19 and 16 Return your textbooks!

Entry Task: Entry Task: June 11June 11thth Review Ch, 15, 19 Review Ch, 15, 19 and 16and 16Return your Return your textbooks!textbooks!

Ch. 15 Ch. 15 & 19& 19SolutionSolutions & s & Acid-Acid-BaseBase

FORGOT THE FORMULAS??

Dilutions: M1V1 = M2V2

1. What is the percent by mass of NaHCO3 in a solution containing 20 grams of NaHCO3 dissolved in 600 ml H2O?

Identify the solute___________ Identify the solvent____________NaHCO3 Water

= % NaHCO320g NaHCO3

600ml water

20g NaHCO3 + 600ml water = 620 ml solution

X 10020g NaHCO3

620ml solution = 3.2% NaHCO3

2. You have 1500.0 grams of bleach solution. The percent by mass of the solute sodium hypochlorite, NaOCl, is 3.62%. How many grams of NaOCl are in the solution? How many grams of solvent are in the solution?

= 3.62% NaOClXg NaOCl1500 g solution

= 3.62 100

Xg NaHCO3

1500 g solution

= 54.3g NaOCl

1500 X 3.2 = 5430 g 100

= 1446 g Solvent1500g solution = solvent + solute (54.3 g)

3. What is the molarity of an aqueous solution containing 40.0g of glucose (C6H12O6) in 1.5L of solution?

40 g of C6H12O6 1 mole of C6H12O6

0.222 moles of C6H12O6

1.5 L solution= 0.148 M

180.16 g C6H12O6

4. Calculate the molarity of 1.60 L of a solution containing 1.55 g of dissolved KBr.

1.55 g of KBr

119.0 g KBr

1 mole of KBr

0.013 moles of KBr

1.60 L solution= 0.00814 M

5. If I add water to 100 ml of a 0.15 M NaOH solution until the final volume is 150 ml, what will the molarity of the the diluted solution?

(0.15 M)(100 ml) = (X M)(150 ml)

15150

= 0.1 M NaOH

6. How much 0.05 M HCl solution can be made by diluting 250 ml of 10 M HCl?

(10 M)(250 ml) = (0.05 M)(X ml)

25000.05

= 50000 ml of HCl

_____1. proton acceptor _____12. metal and nonmetal

_____2. red with pH indicator _____13. positively charged solution

_____3. ionicly bonded _____14. H2SO3

_____4. OH on the back of its formula _____15. covalently bonded

_____5. negatively charged solution _____16. nonmetal-nonmetal

_____6. does not react with metals _____17. Ca(OH)2

_____7. reacts to carbonates _____18. does not react with carbonates

_____8. H in the front of its formula _____19. proton donor

_____9. hydronium ions _____20. pH of 0-6

_____10. reacts with metals _____21. hydroxide ions

_____11. pH 8-14 _____22. purple with pH indicator

BABBBBAAAAB

BAAAABBAABB

Name the following acids and bases

HF Phosphoric acid

H2SO3 Carbonic acid

Zn(OH)2 Barium hydroxide

Hydrofluoric acid H3PO4

sulfurous acid H2CO3

Zinc hydroxide Ba(OH)2

acid base conjugate acidconjugate baseHF + SO3

2– F– + HSO3–

1. If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H2SO4), what is the concentration of the H2SO4 solution?

KOH + H2SO4 K2SO4 + H2O

X Macid 125mlacid = (2) 0.5 Mbase 50mlbase

50125

= 0.4M acid

2 2

Ch. 16Ch. 16EnergyEnergy

Determine the mass of a sample of silver if 705 J of heat are required to raise its temperature from 25oC to 35oC. The

specific heat of silver is 0.235 J/g.oC.

q=cmT• q = 705J• c= 0.235 J/g.oC• m= X g• ΔT= 25-35=10

705 J10 x 0.235

300 g

A 7.5 g nugget of pure lead absorbs 276 J of heat. What was the final temperature of lead if the initial temperature was

25.0˚C? (The specific heat of lead is 0.129 J/g.oC)

q=cmT• q = 276J• c= 0.129 J/g.oC• m= 7.5 g• ΔT= X

276 J7.5 x 0.129

285 °C

Since the temperature started at 25.0°C, the final temperature is 310°C.

Identify which of the following is an endothermic or exothermic process

• EndothermicLiquid gas

Has a positive ΔH value

Solid gas

Liquid solid

• Endothermic

• Endothermic

• Exothermic

2C2H4O(l) + 2H2O(l) → 2C2H6O(l) + O2(g)

C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -685.5 kJ

C2H4O(l) + 2.5O2(g) → 2CO2(g) + 2H2O(l) ΔH = -583.5 kJ

  4CO2(g)  +  6H2O(l) →  2C2H6O(l)  +  6O2(g)   ΔH = 1371 kJ

2C2H4O(l)  +  5O2(g)  →  4CO2(g)  +  4H2O(l)   ΔH = -1167 kJ

2C2H4O(l) + 2H2O(l)  →  2C2H6O(l) +  O2(g) ΔH= 204

CH2O(g) +  H 2(g)   →    CH4O(l)                       ΔH = 65 kJ

N2H4(l) + H2(g) → 2NH3(g)

N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2 (g) ΔH = -37 kJ

N2(g) + 3H2(g) → 2NH 3(g) ΔH = -46 kJ

CH4O(l) → CH2O(g) + H 2(g) ΔH = -65 kJ

N2H4(l)  +  CH4O(l)  →  CH2O(g)  +  N2(g)  +  3H2 (g)    ΔH = -37 kJ

N2(g)  +  3H2(g)  →  2NH 3(g)                             ΔH = -46 kJ

N2H4(l)  +  H2(g)  →  2NH3(g) ΔH= -18kJ