CHEM 1201 : Exam 1

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azuremammoth471

15 Mar 2019

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It decomposes to form an off-white solid b and a gas c. the gas has the same properties as the product obtained when carbon is burned in excess oxygen. Cah2 + 2h2o ca(oh)2 + 2h2: 76 g cah2, 38 g cah2, 103 g cah2, 15 g cah2, 42 g cah2. The molecular formula for aspartame (nutrasweet) is c14h18n2o5. (mw = 294 g/mol) Determine the empirical formula of the compound with the mass composition. Hclo: 4. 38 1020molecules, 2. 05 1018 molecules, 1. 99 1022 molecules, 6. 22 1021 molecules, 5. 73 1015 molecules. 24. 5% na, 14. 9% si, 60. 6% f: na2sif6, nasi2f5, na3sif3, na2si2f3, nasi2f4. Choose the correct balanced net ionic reaction when a lead(ii) acetate solution reacts. Specify which chemical species are present when nh4c2h3o2 dissolves in water. Calculate the concentration of the ions in a solution prepared by dissolving 8. 21 g of: n3- and c2h7o23, nh3 and c2h4o2.

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Part 1 Boyle’s Law P1 V1 = P2 V2 when Temperature is constant; being 1 “initial” and 2 “final” A container of oxygen gas has a volume of 150 mL when its pressure is 0.947 atm. If the temperature remains constant. What will the volume of the gas be at a pressure of 0.987 atm? Using Boyle’s Law solve for V2 of final volume using P1 V1 = P2 V2 Match the correspondent given data, unknown data and resulted in data after solved

___1. Initial Volume or V1

___2. Final Pressure or P2

___3. Initial pressure or P1

___4. Final V2

___5 Temperature

A. does not change, remains constant

B. 144 mL of oxygen

C. 0.947 atm

D. 150 mL of oxygen

E. 130 mL of oxygen

F. 0.987 atm

Part 2 Charles’s Law V1/T1=V2/T2 when the Pressure is constant; being 1 “initial” and 2 “final A container with hydrogen gas has a volume of 349 mL, at a temperature of 22.0C (Change to K), What volume will the gas occupy at 50C(Change to K), if the pressure remains constant?

____6. Final Volume of gas

____7 Initial Temperature gas

____8. Final Temperature gas

____9. Initial Volume gas

____10. The pressure of hydrogen

G. It’s constant

H. 295 K

I. 349 mL

J. 382 mL

K. 323 K

L. 50 C

Part 3 Gay-Lussac Law P1/T1=P2 when the volume (V) is constant A cylinder of a gas has a pressure of 4.40 atm at 25C (Change to K). At what temperature in Celsius will it reach a pressure of 6.50 atm C= K-273

____11. Final Pressure of gas (P1)

____12. The initial temperature in K (T1)

____13. The initial Pressure of gas (P2)

____14. The volume of the gas

_____15. The final temperature of the gas

M. 6.50 atm

N. 298 K

O. 4.40 atm

P. 167 C

Q . 283 K

R. is constant

Part 4 Combined Gas Law P1 V1/ T1 = P2 V2/ T2 OR MORE EASY TO SOLVE P1 V1 T2= P2 V2 T1 A St. Valentine Helium-filled balloon for your love, has a volume of 50.0 mL at 25 C (change to K) and 1.08 atm. What volume(V2) will it have at 0.855 atm and 10.0C (change to K). K=C + 273

_____16. The Final Temperature in K (T1)

_____17. The initial Volume in mL (V1)

_____18. The final Pressure in atm (P2)

_____19. The initial temperature in C (T1)

_____20. The final volume (V2)

S. 298k

T. 50.0 mL

U. 0.855 atm

V. 25 C

W. 283 K

X. 60.00 mL

Part 5 Answer with the letter in the blank. A. Boyles’s Law B. Charles’s Law C. Gay-Lussac’s Law D. Combined Gas Law E. Avogadro’s Law

___21. One mol of any gas at Standard Temperature and Pressure contains 6.03 X10^23 molecules and occupy a volume of 22.4L

____22. The higher the pressure the lower the volume when the temperature is constant.

____23. The higher the temperature the higher the volume when the pressure is constant.

____24 The higher the temperature the higher is the pressure when the volume is constant.

____25. There is a relation between pressure, volume, and temperature of a fixed amount (moles)of gas

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CHEM 1201 : Exam 1

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