1. Dr. Laila Mohammed Al-Harbi Assistant professor Contact Info: Lalhrbi@kau.edu.sa Lalhrbi@kau.edu.sa Web Site: http://lalhrbi.kau.edu.sa

2.  Locations: ◦ Science tower 07 room 173 first floor ◦ phone 6400000 ext. 23024 ◦ email Lalhrbi@kau.edu.saLalhrbi@kau.edu.sa ◦ web site: http://lalhrbi.kau.edu.sahttp://lalhrbi.kau.edu.sa  Exam schedule:  1 st exam : from lecture 1-11 ( Chapters 1-4) = 30 marks  2 nd exam: from lecture 12-24 (Chapters 5,7-9) = 30 marks  Final exam: from lecture 1-33 = 40 marks  (Chapters 1,2,3,4,5,7,8,914,15,24&25)  Dr. LAILA AL-HARBI

3. CHEM 110 1Course No.Course Title No. of Units Pre-requisites Th.Pr.Credit Chem 110General Chemistry I3-3- Course Objectives: The course aims to introduce students to basic knowledge and principle in chemistry. Course Description : It provides an introduction to the general principles of chemistry for students planning a professional career in chemistry, a related science, the health professions, or engineering. By the end of this course the student will be able to understand the following: Significant figures, scientific notation and units, stoichiometry, atomic structure & periodic table, chemical bonding, gases, ionic equilibrium, basic principles of organic and basic principles of biochemistry. Main text books:  Chemistry, by Chang, 10 th. ed., 2007, McGraw-Hill.  Chemistry, by Steven S. Zumdahl, 6 th ed., Houghton Mifflin College Div. Subsidiary books : Chemistry, by Mortimer, 6 th ed., Wadsworth Inc. Dr. LAILA AL-HARBI

4. CHEM 110 Main text book :  Chemistry, by Chang, 10 th. ed., 2007, McGraw- Hill. Dr. LAILA AL-HARBI

5.  المذاكرة أول باول  عدم التغيب عن المحاضرات  مذاكرة الكتاب الدراسي و المحاضرات  حل بنك الاسئلة... حيث يحوي على مجموعة مختلفة من الاسئلة بافكار متعدده... تساعدك على التدرب  التدرب على الاسئلة في الموقع التفاعلي حيث أن الطريقة شبيهة تماما بالاختبار الالكتروني  http://prod.kau.edu.sa/faculties/science/website2/index.aspx http://prod.kau.edu.sa/faculties/science/website2/index.aspx  و أخيرا... التوفيق بيد الله سبحانه و تعالي Dr. LAILA AL-HARBI

6.  Measurement  SI units  Mass and weight  Volume  Temperature scales Dr. LAILA AL-HARBI

7.  By the end of this chapter you should:  Know the 7 SI basic units and their prefixes.  Be able to convert from one unit to other.  Know to derive units from the 7 SI basic units.  Common units (L & mL)  Know the temperature Scales Dr. LAILA AL-HARBI

8.  Chemistry is the study of matter and the changes it undergoes  There are three states of matter solid liquid gas Dr. LAILA AL-HARBI

9.  used for commerce and science around the world Dr. LAILA AL-HARBI

10. 10 The SI unit of mass is (a) The pound (b) The gram (c) The kilogram (d) The mole Examples The Kg is the SI unit of (a) length (b) mass (c) temperature (d) current

11. 1m Dr. LAILA AL-HARBI

12.  A ) use numerical line خط الاعداد  B) use transfer factor معامل التحويل للتحويل من الوحده الاساسية و مضاعفتها أو المشتقات يمكننا استخدام خط الاعداد أو استخدام معامل التحويل (بضرب الطريفين في الوسطين ) و لكن لا تخلطي بين الطريقتيين كما أن الطريقتين يعطوا نفس النتيجة اذا استخدموابطريقة صحيحة. Dr. LAILA AL-HARBI

13. للتحويل من الوحده الاساسية و مضاعفتها أو المشتقات يمكننا استخدام خط الاعداد أو استخدام معامل التحويل (بضرب الطريفين في الوسطين ) و لكن لا تخلطي بين الطريقتيين كما أن الطريقتين يعطوا نفس النتيجة اذا تم استخدامها بنفس بطريقة صحيحة. TGM KSI d c m µ n p 10 3 10 10 3 10 12 10 9 10 6 10 3 10 12 10 9 10 6 10 3 10 2 10 معامل التحويل بهذا الاتجاه تزيد الوحده وحدة صغيرة وحدة كبيرة للتحويل من صغير الى كبير نقسم ÷, للتحويل من كبير الى صغير نضرب ×

14. Dr. LAILA AL-HARBI

15. examples The base unit is meter (m) The base unit is gram (g)The base unit is liter (L) استخدام البادئات (مضاعفات أو أجزاء) هو من أجل التبسيط والاختصار للكميات المقاسة

16. Tera (T)= 10 12 Gega (G)= 10 9 Mega (M)= 10 6 kilo (k)= 10 3 hecto (h)= 10 2 deca (da)= 10 1 unit (meter-m), (gram-g), (liter-L) dece (d)= 10 -1 centi (c)= 10 -2 mili (m)= 10 -3 micro (  )= 10 -6 nano (n)= 10 -9 pico (p)= 10 -12 femto (f)= 10 -15 Bigger unit smaller unit atto (a)= 10 -18 Exa (E)= 10 18 الأجـــزاء المضاعفات

17. 1- التحويل بين وحدات أكبر من الوحدة الأساسية (المضاعفات) (نأخذ فرق الأسس المطلقة بين الوحدات المراد التحويل بينها) 2- التحويل بين وحدات أصغر من الوحدة الأساسية (الأجزاء) (نأخذ فرق الأسس المطلقة بين الوحدات المراد التحويل بينها) dece (10 -1 ) nano (10 -9 ) 3- التحويل بين المضاعفات والأجزاء مروراً بالوحدة الأساسية (نجمع الأسس المطلقة للوحدات المراد التحويل بينها) Mega (10 6 ) nano (10 -9 ) في حالة التحويل من وحدة كبيرة إلى وحدة صغيرة نضرب في معامل بأس موجب أما في حالة التحويل من وحدة صغيرة إلى وحدة كبيرة نضرب في معامل بأس سالب ملحوظة مهمة

18. Dr. LAILA AL-HARBI I. Convert 134 pm to m? II. How many meters are in 134 pm? للتحويل من صغير الى كبير نقسم ÷ 1 pm = 1×10 -12 m 134 pm = ?? m الحل باستخدام خط الاعداد الحل باستخدام الجدول طرفين في وسطين 134 ×1×10 -12 m = 1.34×10 -10 m

19.  The SI prefixes giga and micro represent, respectively:  A.10 -9 and 10 -6.  B.10 6 and 10 -3.  C.10 3 and 10 -3.  D.10 9 and 10 -6.  The SI unit of mass is  (a). The pound  (b). The gram  (c). The kilogram  (d). The mole. Dr. LAILA AL-HARBI

20. How many microseconds are in a second? (a). 1 x 10 -1 (b). 1 x 10 -6 (c). 1 x 10 -15 (d). 1 x 10 6 Explanation: Since the prefix micro means 1 x 10 -6, there will be 1 x 10 + 6 microseconds in one second. Dr. LAILA AL-HARBI 1 µs= 1 x 10 -6 s x µs = 1 s s µs 1 10 6 x 1 x 10 6 µs

21.  Which of the following is the smallest distance?  (a) 21 m → 21m  (b) 2.1 x 10 2 cm → 2.1m  (c) 21 mm → 0.021 m  (d) 2.1 x 10 4 pm → 2.1 x 10 -8 m Put all of them in the same unit  Explanation: Even though 2.1 x 10 4 is the largest number in this question, the units of pm (picometers) are the smallest units here, making it the smallest distance. Dr. LAILA AL-HARBI

22.  A) 6.0 km is how many micrometers?  Solution 1 1 km = 10 3 m 6 km = x = 6 × 10 3 m 1µm = 1× 10 -6 m x = 6 × 10 3 m x = 6 × 10 9 µm Dr. LAILA AL-HARBI Explanation: convert first to meter then from meter to micro ( two steps solution ) km µm × 6 10 9 6 × 10 9

23.  Example  The diameter of an atom is approximately 1  10 -7 mm. What is this diameter when expressed in nanometers?  A.1  10 -18 nm  B.1  10 -15 nm  C.1  10 -9 nm  D.1  10 -1 nm  = 1× 10 -7 × 1 × 10 6 =  1 × 10 -1 nm = 0.1 nm  Example  Which of these quantities represents the largest mass?  A.2.0  10 2 mg  B.0.0010 kg  C.1.0  10 5  g  D.2.0  10 2 cg  Put all of them in the same unit Dr. LAILA AL-HARBI A) 0.2 g B)1 g C) 0.1 g D) 2 g

24.  are defined in terms of the seven base quantities via a system of quantity equations.  The SI derived units for these derived quantities are obtained from these equations and the seven SI base units. For example  Area = width x length  Unit of width = m  Unit of length = m  Unit of Area = m× m = m 2 Dr. LAILA AL-HARBI Treat units like numbers

25. Volume – Volume = width × length × heights = m × m × m = m 3 SI derived unit for volume is cubic meter (m 3 ) Common unit of volume is liter (L) and milliliter (ml) The relation ship between liter (L) and ml (1L= 1000mL) The relation ship between liter (L) and metric system 1 L = 1 dm 3 The relation ship between milliliter (ml) and metric system 1 mL = 1 cm 3 1 cm 3 = (1 x 10 -2 m) 3 = 1 x 10 -6 m 3 1 dm 3 = (1 x 10 -1 m) 3 = 1 x 10 -3 m 3 Dr. LAILA AL-HARBI

26. Example (1) Example (2)  How many liters are in 25 dm 3 ?  Science 1L = 1 dm 3  25 dm 3 = 25L  How many milliliters are in 32 cm 3 ?  Science 1mL = 1 cm 3  32 cm 3 = 32 mL Dr. LAILA AL-HARBI Example (3)  How many liters are in 250 cm 3 ?  Science 1L = 1 dm 3  and 1mL = 1 cm 3  250 cm 3 = 250 mL  L → mL  250/1000 = 0.25 L

27.  How many cubic centimeters are there in exactly one cubic meter?  A.1  10 -6 cm 3  B.1  10 -3 cm 3  C.1  10 -2 cm 3  D.1  10 6 cm 3  Solution  (1m ) 3 = (cm ) 3  1m 3 = (1  10 2 ) 3 cm 3  1m 3 = 1  10 6 cm 3 Dr. LAILA AL-HARBI  The diameter of an atom is approximately 1  10 -7 mm. What is this diameter when expressed in nanometers?  A.1  10 -18 nm  B.1  10 -15 nm  C.1  10 -9 nm  D.1  10 -1 nm

28.  Mass is the measure of the amount of matter in an object. SI unit of mass is the kilogram (kg) 1 kg = 1000 g = 1 x 10 3 g  Weight is the measurement of the pull of gravity on an object.  The Mass of an object doesn't change when an object's location changes. Weight, on the other hand does change with location.  Chemist are interested primarily in mass weight = c x mass The weight of man on earth is 50 pounds. is 8.25 pounds on moon Dr. LAILA AL-HARBI

29. Density is defined as the amount of matter in a given amount of space. is kg/m 3  SI derived unit for density is kg/m 3  common units of density are g/mL, g/L  Density decrease with temperature (g/ml )g/cm 3 for liquid and solids g/L = 0.001g/ml for gases  Because density of gases are very low The density of copper is 8.94 g/cm 3. density = mass volume Dr. LAILA AL-HARBI

30. 30 Given m & V ? d d = m/V Given m & d ? V V = m /d Given V & d ? m m = d x V

31.  A piece of Gold metal has a volume of 15.6 cm 3, with a mass of 301 g What is its density?  A piece of platinum metal with a density of 21.5 g/cm 3 has a volume of 4.49 cm 3. What is its mass? d = m V m = d x V = 21.5 g/cm 3 x 4.49 cm 3 = 96.5 g d = m V 301 g/ 15.6 cm 3 = 19.3 g/ cm 3 Dr. LAILA AL-HARBI

32.  The density of mercury is 13.6 g/mL has a volume of 5.50 mL. What is its mass?  The density of sulfuric acid is 1.41 g/mL has a volume of 242 mL. What is its mass? d = m V m = d x V = 1.41 g/mL x 242 mL = 341.22 g d = m V m = d x V = 13.6 g/mL x 5.50 mL = 74.8 g Dr. LAILA AL-HARBI

33.  Fahrenheit °F →°F = [ (9/5) × °C] + 32  Celsius °C → °C = (5/9) (°F - 32)  Kelvin ° K → ° K = °C + 273.15 Dr. LAILA AL-HARBI

34. 34 Temperature Units Conversion 1.Degrees Celsius 0 C: Scale 0 → 100 Thus: 100 divisions or 100 degrees 2. Kelvin K: Scale 273 → 373 Thus: 100 divisions or 100 degrees 1K = 1 C 3. Degrees Fahrenheit 0 F : Scale from 32 → 212 Thus: 180 divisions or 180 degrees Thus: the size of degree in 0 F scale is only 100/180 or 5/9 of a degree on the 0 C scale 1 0 F = (5/9) 1 0 C

35.  K → °C or °C → K K = °C + 273.15 °C = K - 273.15  °F → °C or °C → °F °F = [ (9/5) × °C] + 32 °C = (5/9) (°F - 32)  No relatationship between K → °F or °F → K Thus: K → 0 C → 0 F or 0 F → 0 C → K Temperature Units Conversion

36.  Convert 224 0 C to degrees Fahrenheit?  °F = (9 0 F /5 0 C) × °C + 32  [°F = (9 0 F /5 0 C) × 224 °C] + 32 0 C = 435 0 F  Convert -452 0 F to degrees Celsius.  °C = (5 0 C /9 0 F) (°F - 32 0 F)  °C = (5 0 C /9 0 F) (-452 °F - 32 0 F) = -269 0 C  Convert -38.9 0 C to degrees Kelvin..  ° K = [-38.9 °C + 273.15 °C ] × 1 K/ 1 0 C = 234.3 K Dr. LAILA AL-HARBI

37.  Ammonia boils at -33.4  C. What temperature is this in  F?  A.-60.1  F  B.-92.1  F  C.-28.1  F  D.+13.5  F F = (9 0 F /5 0 C) × °C + 32 [°F = (9 0 F /5 0 C) × -33.4 °C] + 32 0 C = - 28.1 0 F Dr. LAILA AL-HARBI

38.  Problems  1.17 - 1.18 – 1.19 – 1.22 - 1.23 – 1.25 – 1.26 Dr.S. Al- Mhy awi