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Principles of Chemistry I

Principles of Chemistry II

MODULE 4 UNIT 2 Assignment Part A Write the formulas for compounds 1 to 15 and the names of the compounds represented by formulas 16 to 30. 1. Silver nitrate AgNO 3 2. Tin (IV) sulphate SnSO 4 3. Iron (III) nitrate Fe(NO 3 ) 3 4. Copper (II) carbonate CuCO 3 5. Lead (II) phosphate Pb 3 (PO 4 ) 2 6. Mercury (I) nitrate Hg(NO 3 ) 2 7. Sodium chlorate NaClO 3 8. Caesium phosphate Cu 3 P 2 9. Calcium acetate Ca(C 2 H 3 O 2 ) 2 10. Zinc nitrate Zn(NO 3 ) 2 11. Aluminium sulphate Al 2 (SO 4 ) 3 12. Lithium chlorate LiClO 3 13. Copper (II) phosphate Cu 3 (PO 4 ) 2 14. Mercury (I) chlorate Hg 2 (ClO 3 ) 2 15. Sodium carbonate NaHCO 3 16. CuSo 4 Copper Sulfate 17. Mg(NO 3 ) 2 Magnesium Nitrate 18. CaSO 4 Calcium Sulfate 19. Zn(C 2 H 3 O 2 ) 2 Zinc Acetate 20. AlPO 4 Aluminium Phosphate 21. RbNO 3 Rubidium Nitrate 22. HgCO 3 Mercury Carbonate 23. K 2 SO 4 Potassiun sulfate 24. Sn 3 (PO 4 ) 4 Tin (IV) phosphate 25. CaCO 3 Calcium Carbonate 26. Li 2 SO 4 Lithium sulfate 27. CuNO 3 Copper Nitrate 28. Ag 2 SO 4 Silver sulfate 29. BaSo 4 Barium sulfate 30. K 3 PO 4 Potassium phosphate Total 30 points Submission Instructions Submit your assignment using the drop box on the Assignment page for this unit.

CHEM | 1152 | Survey of Chemistry II | homework | MODULE 4 UNIT 2 Assignment Par

MODULE 2 UNIT 1 Assignment Part E Submission Instructions Submit your assignment using the drop box on the Assignment page for this unit. At the end of the document, include a list of relevant sources of information you used to complete this assignment. 1. Refer to the detailed periodic table included with this module and indicate the trends showing along the rows and columns with regard to the following properties: a) atomic size along groups 2 marks As you go along groups from top to bottom, the atomic size increases. b) atomic size along periods 2 marks As you move along periods left to right, the atomic size gradually decreases How are atomic sizes related to the number of protons / electrons in atoms? 2 marks The atomic sizes are related to the number of protons and electrons in the atoms because as you move to the right, the atomic number increases which in relation’s means that the number of protons increases, as well as the number of electrons. Total 6 marks https:// www.lenntech.com/periodic/periodic-chart.htm

CHEM | 1152 | Survey of Chemistry II | homework | MODULE 2 UNIT 1 Assignment Par

1 MODULE 9 UNIT 1 Assignment Part B 1. What is an alkene? Give the general formula for alkenes. 2 mark Alkenes are hydrocarbons that have one or more double bonds between carbon atoms. The formula of chain alkenes is C n H 2n 2. Name a diﬀerence between a saturated and an unsaturated hydrocarbon. 1 mark A saturated hydrocarbon contains only single bonds as all carbon atoms are saturated with hydrogen atoms. In unsaturated molecules, there is a double bond between two carbon atoms. If there is more than one double bond, they are known as polyunsaturated hydrocarbons. 3. What type of organic compound is ? Name It. 3 marks 2-methylpentane 4. What type of hydrocarbon is ? Name it. 2 marks 2-butane 5. Write the structural formula and the name of the organic product that would result from the hydrogenation (addition of hydrogen) reaction of propene. 3 marks CH3CH2CH2 Propane 2 6. Write the name and the structural formula of the organic product that would result from the reaction below. Don’t forget the Markovnikov’s rule. CH 3 CH 2 CH=CH 2 + HBr->. 3 marks CH3CH2CHBrCH3 2-bromo-butane 7. Identify the type of organic compound resulted after adding a water molecule to 2-hexene. Then, write its structural formula. 3 marks It would be an alcohol because in the industrial process of the production of ethanol, you add steam (water as gas) to ethane. The formula will be C6H13OH – (CH3CH2CH2CH2CH2CHOH) 8. Write the structural formula of organic product that would result from the reaction: CH 3 —CH=CH 2 + H 2 O->. 2 marks CH3-CH=CH2+H2O-  CH3-CH(OH)- CH3 CH3-CH(OH)-CH3 Organic product is 2-propanol 9. What do the preﬁxes ortho-, meta-, and para- refer to in terms of the relative location of substituents in a disubstituted aromatic benzene? 2 marks The preﬁxes, ortho, meta, and para is to help distinguish di- substituded benzene dirivatives as well to denote speciﬁc relative locations of the substituents on an aromatic ring. 10. Name the aromatic compound . 2 marks 1,3 dimethylbenzene Submission Instructions Submit your assignment using the drop box on the Assignment page for this unit. 3 https://olmoodle.ontariolearn.com/mod/resource/view.php?id=349038 4

CHEM | 1152 | Survey of Chemistry II | homework | 1 MODULE 9 UNIT 1 Assignment P

1 MODULE 5 UNIT 1 Assignment Part D 1. A compound whose empirical formula is CH 3 has a molar mass of 30 g/mol. What is its molecular formula? (3 points) Molar Mass = 30 g/mol= 2 Mass of empirical formula 15 g/mol The molecular formula is 2 x CH3= C2H6. 2. Find the molecular formula for each of the following compounds: (6 points) a) 30.5% nitrogen; 69.5% oxygen; molar mass of 46 g/mol Number of moles of N = 30.5 g = 2.18 mol 14.0 g/mol Number of moles of O =69.5 g = 4.34 mol 16.0 g/mol Molar Ratio: N = 2.18 mol / 2.18 mol= 1 O = 4.34 mol/2.18 mol= 2 The empirical formula is NO2. Molar Mass = 46 g/mol= 1 Mass of empirical formula 46 g/mol The molecular formula is NO2. b) 2.67 g of carbon; 0.22 g of hydrogen; 7.11 g of oxygen; molar mass of 90 g/mol Hint: Start by calculating the empirical formula. Number of moles of C = 2.67 g / 12.0 g/mol = 0.22 mol Number of moles of H = 0.22 g / 1.0 g/mol = 0.22 mol Number of moles of O = 7.11 g / 16.0 g/mol= 2 0.44 mol Molar Ratio: C = 0.22 mol / 0.22 mol= 1 H = 0.22 mol / 0.22 mol= 1 O = 0.44 mol / 0.22 mol= 2 The empirical formula is CHO2 3 Molar Mass = 90 g/mol= 2 Mass of empirical formula 45 g/mol The molecular formula is C2H2O4. 3. A gas sample of 17.0 g is constituted of 15.8 g of carbon, the rest being hydrogen. Given that the sample’s volume is 4.9 L at NTP conditions, calculate the molecular formula for this compound. (5 points) 17g gas sample -15.8g carbon = 1.2g hydrogen C: 12.01g/mol. 15.8g/12.01g/mol = 1.3 moles H: 1.01g/mol. 1.2g/1.01g/mol = 1.2 moles 1.3/1.2 = 1.1 The empirical formula is CH Empirical mass is 12.01g + 1.01g = 13.02g 4.9L/22.4L/mol = 0.22 mol Molar mass is 17g/0.22mole = 77.3g/mol77.3g/13.02g = 5.9 = 6 6 x CH = (CH) 6 The molecular formula is (CH) 6 4. A compound is constituted by mass of 22.6% phosphorus, the rest being chlorine. Given that a volume of 307 mL of the gas has a mass of 1.55 g at 25 °C and 90 kPa, ﬁnd the molecular formula for this compound. (5 points) 1.55g / 100% = 0.016 x 22.6% = 0.35g of phosphorus 1.55g / 100% = 0.016 x 77.4% = 1.20g of chlorine P: 30.97g/mol. 0.35g/30.97g/mol. =0.01mol Cl: 35.45g/mol. 1.20g/35.45g/mol. = 0.03mol 0.03/0.01 = 3 to 1 ratio The empirical formula is PCl 3 Empirical mass is 30.97g + 35.45 (x3= 106.35g) = 137.32g0.307L x 298/273 (=1.09) x 90/101 (=0.89) = 0.298L 0.298L/22.4L = 0.0133 = 1.33 x 10-2mole Molar mass = 1.55g/1.33 x 10-2mole = 116.54g/mol / 137.32g = 0.84 = 1 x PCl 3 The molecular formula is PCl 3 4 5. A compound is constituted by mass of 52.1% carbon and 13.1% hydrogen, the rest being oxygen. Given that a volume of 5.33 L of this gas has a mass of 11.5 g at 13 °C and 110 kPa, ﬁnd the molecular formula for this compound. (5 points) 100-52.1-13.1 = 34.8 11.5g/100% x52.1= 5.99g C

CHEM | 1152 | Survey of Chemistry II | homework | 1 MODULE 5 UNIT 1 Assignment P

O O Cl Cl I Cl Cl C O O 1 7. Which of the following statements is true regarding the Lewis structure of the carbonate anion (CO 3 2– )? A. It contains one C–O single bond and two C=O double bonds. B. It contains two C–O single bonds and one C=O double bond. C. It contains three C=O double bonds. D. It contains two C–O single bonds and one C ≡ O triple bond. E. It contains one O

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