qaterly 3 review key

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QATERLY 3 REVIEW- key Chapter 12 – Stoichiometry 6. a. 537g Fe b. 93.1g Fe 7. 14.7g of Ba 3 (PO 4 ) 2 ; Limiting reactant- Ba(NO 3 ) 2 ; Excess left over- 9.5g Na 3 PO 4 ; 83.7% yield Chapter 13 – Solids and Liquids 1. dipole-dipole, dispersion 2. dispersion 3. dispersion 4. H-bonds, dipole-dipole, dispersion 5. Solids- definite shape/volume, organized & close, vibrate around fixed point. Liquids- definite volume, shape of container, unorganized yet close, move past each other. 6. network 7. metals 8. molecular 9. amorphous 10. ionic 11. strong IMF = low volatility, low vapor pressure, high bp 12. 28.0 g/mol Chapter 14 – Gases 13. 1180 mmHg 14. 180. kPa 15. 493°C 16. 80. L 17. 440. L 18. 43.6 g/mol 19. 7.94 L 20. 93.2 K Chapter 15 – Solutions 21. Unsaturated-dissolves, saturated-sinks to bottom, supersaturated-excess crystallizes 22. temp↑ solubility solids ↑ solubility gases pressure ↑ solubility gases ↑ 23. soluble 25. soluble 24. insoluble 26. insoluble 27. a. 40 grams c. CO 2 downwards b. ~7 grams 28. 9.00 g 29. 4.2 mL 30. Molarity= mol solute/L of solution Molality = mol solute/kg of solvent 31. AlCl 3 32. Fp = -4.76°C, bp = 101.31°C

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Qaterly 3 Review Key

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QATERLY 3 REVIEW- key

Chapter 12 Stoichiometry

6. a. 537g Fe b. 93.1g Fe 7. 14.7g of Ba3(PO4)2; Limiting reactant- Ba(NO3)2; Excess left over- 9.5g Na3PO4; 83.7% yieldChapter 13 Solids and Liquids

1. dipole-dipole, dispersion2. dispersion3. dispersion4. H-bonds, dipole-dipole, dispersion5. Solids- definite shape/volume, organized & close, vibrate around fixed point. Liquids- definite volume, shape of container, unorganized yet close, move past each other.6. network7. metals8. molecular9. amorphous10. ionic11. strong IMF = low volatility, low vapor pressure, high bp12. 28.0 g/mol

Chapter 14 Gases13. 14. 1180 mmHg15. 180. kPa16. 493C17. 80. L18. 440. L19. 43.6 g/mol20. 7.94 L21. 93.2 K

Chapter 15 Solutions

22. Unsaturated-dissolves, saturated-sinks to bottom, supersaturated-excess crystallizes23. temp solubility solids solubility gases pressure solubility gases 24. soluble25. soluble25. insoluble26. insoluble27. a. 40 gramsc. CO2 downwardsb. ~7 grams28. 9.00 g29. 4.2 mL30. Molarity= mol solute/L of solutionMolality = mol solute/kg of solvent31. AlCl332. Fp = -4.76C, bp = 101.31C

EXTRA PRACTICE PROBLEMS key1. What relationships can be determined from a balanced chemical equation?Mole to mole1. Explain why mole ratios are central to stoichiometric calculations.Since atoms and molecules have different masses, we need to compare moles to have the correct number of particles.

3. How many grams of silver phosphate are produced? 3AgNO3 + Na3PO4 Ag3PO4 + 3NaNO3

200.0g AgNO3 x 1 mole AgNO3 x 1 mole Ag3PO4 x 418.58g Ag3PO4 = 164.3g Ag3PO4 169.88g AgNO3 3 mole AgNO3 1 mole Ag3PO4

200.0g AgNO3 x 1 mole AgNO3 x 1 mole Ag3PO4 x 418.58g Ag3PO4 = 164.3g Ag3PO4 169.88g AgNO3 3 mole AgNO3 1 mole Ag3PO4

200.0g Na3PO4 x 1 mole Na3PO4 x 1 mole Ag3PO4 x 418.58g Ag3PO4 = 510.6g Ag3PO4 163.94g Na3PO4 1 mole Na3PO4 1 mole Ag3PO4

4a. What volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and 75.0 g of HCl? Mg + 2HCl MgCl2 + H2

50.0g Mg x 1 mole Mg x 1 mole H2 x 22.4L H2 = 46.1 L H2 24.31g Mg 1 mole Mg 1 mole H2

75.0g HCl x 1 mole HCl x 1 mole H2 x 22.4L H2 = 23.0 L H2 36.46g HCl 2 mole HCl 1 mole H2 so HCl is LR

75.0g HCl x 1 mole HCl x 1 mole Mg x 24.31 Mg = 25.0g Mg reacted 36.46g HCl 2 mole HCl 1 mole Mg so 50.0g 25.0g=25.0g

b. How much excess reactant is left over?

5. A balloon is filled with 3.0 L of helium at 1 atm. What is the volume when the balloon rises to an altitude where the pressure is only 0.25 atm? (assume temperature remains constant.)P1 V1 = P2 V2 (1.0 atm)(3.0L)=(0.25 atm)(X)

= 12L

6. The gas left in a used aerosol can is at a pressure of 1 atm at 27C. If this can is thrown into a fire, what is the internal pressure of the gas when its temperature reaches 927C?P1 = P2 1 atm = P2 T1 T2 300K 1200K

= 4 atm

7. 5.2 L of a gas is at STP. Find the new volume when the temperature rises to 38C and the pressure drops to 600 mmHg.P1V1 = P2V2 T1 T2 P1V1 T2 = P2V2 T1 Solve for the variableV2 = P1V1 T2 solve P2T1 V2 = 7.5L

8. Pressure and volume of a gas at constant temperature are __directly____ proportional. Which scientist said this? ___Boyle_________

9. Volume and temperature of a gas at constant pressure are ___ directly ______ proportional. Which scientist said this? _ Charles __________

10. Volume and number of gas particles at constant pressure and temperature are ___ directly ______ proportional. Avagadro

11. If two containers are at the same temperature and pressure and their volumes are equal, they must contain the same _________# of particles____. Which scientist said this? ________Avagadro______

12. Which travels faster CO2 or O2? How much faster? O2, it travels 1.17x faster

Gas A is faster so we know it is lighter. The lighter molar mass goes on the denominator so that rate comparison is >1

13. Gas A travels 4 times faster than Gas B. If the molar mass of Gas B is 80 g/mole, find the molar mass of Gas A.

square both sides. 80/X=16 X=80/16 X=5 g/mol

14. a. 100C b. gas c. vaporization

15. Ethylen glycol

16. At 1.0 atm , 0.75g of gas dissolves in 1L at 55 degrees. How much of the gas will dissolve in 1L at 4.0 atm at the same temperature? Who came up with this law?___Henry_______

= 3g/LS1 = S2 0.75g/L = X P1 P2 1 atm 4 atm

17. What volume of 3.70 M solution can be prepared using 89.5 grams of sulfuric acid?

3.70M = 0.913moles H2SO4 89.5 g H2SO4 x 1 mole H2SO4 = moles 98 g H2SO4

X L solution =0.246L

18. What mass of magnesium chloride is required to prepare 2.15 liters of 6.50 M solution?

6.50M = X moles MgCl2 X = 14.0 moles MgCl2 2.15 L solution 14.0mole x 95.31g MgCl2 = 1330g (SF) 1 mole MgCl2

19. How would you prepare 2250. mL of a 3.250 M solution from a 16.00 M HNO3 stock solution?M1V1 = M2V2 (16.00 M)(X mL) = (3.250 M)( 2250. mL) dilution!!!!!!!!!!!Take 457.0mL of the stock solution (stronger) and add 1793 mL of water

20. If you mixed the following solutions, what would be the resulting molarity?525 mL of 2.1M HCl Molarity x Volume(L) = moles (2.1M)(0.525L) = 1.1 moles HCl(4.8M)(0.375L) = 1.8moles HCl(7.6M)(0.295L) = 2.2moles HCl 5.1moles total

375 mL of 4.8 M HCl 5.1 moles HCl (total) = 4.3M 1.195L (total)

295 mL of 7.6 M HCl 1195ml

21. Circle all solutes that an i factor of 1?a. C6H12O6b. CF4c. NH4Cl (i=2)d. NH3

22. Why is calcium chloride more effective as rock salt in the winter than sodium chloride? (HINT: what is the i factor for each?) calcium chloride (CaCl2) has an i of 3. NaCl has an i of 2. Calcium chloride will lower the freezing point of water more.