CHEM 105N 1st EditionExam # 1 Study Guide Chapters: 1 - 3Practice Problems and Reminders: For each of the following, indicate whether the unitdescribes length, mass, or volume.________ A. A 10.0 lb bag of tomatoes is 4.5 kg.________ B. A person is 2.0 m tall.________ C. A medication contains 0.50 g of aspirin.________ D. A bottle contains 1.5 L of water.ANSWERS:A. MassB. LengthC. MassD. VolumeIdentify the measurement that has an SI unit.1. John’s height is _____.A. 1.5 yd B. 6 ft C. 2.1 m2. The race was won in _____.A. 19.6 s B. 14.2 min C. 3.5 h3. The mass of a lemon is _____.A. 12 oz B. 0.145 kg C. 0.62 lb4. The temperature is _____.A. 85°C B. 255 K C. 45°FANSWERS:1) C2) A3) B4) B- Standard: 128000 m Scientific Notation: 1.28 x 10^5 m- Standard: .00003 km Scientific Notation: 3 x 10^-5 km- Scientific Notation: 7.2 x 10^-3 g Standard: .0072 g- All non-zero numbers are SIGNIFICANT-Ex: 38.15 cm = 4 Sig Figs 1.88 kg = 3 Sig Figs- Sandwiched Zeros are SIGNIFICANT-Ex: 70.2 = 3 Sig Figs 12003 = 5 Sig Figs .0708 = 3 Sig Figs .390602 = 6 Sig Figs- Trailing Zeros (those that follow non-zero numbers without decimal points) are NOT SIGNIFICANT-Ex: 35000 = 2 Sig Figs 700 = 1 Sig Fig 6800500 = 5 Sig Figs 49000. = 5 Sig Figs- Leading Zeros are NEVER SIGNIFICANT-Ex: .004 = 1 Sig Fig.0431 = 3 Sig Figs- In scientific notation, all numbers in the coefficient are significant-Ex: 3 x 10^4 = 1 Sig Fig 3.0 x 10^4 = 2 Sig Figs 3.00 x 10^4 = 3 Sig Figs- If a question requires a certain number of sig figs, significant zeros may be added-Ex: Three Sig Figs are allowed: 4 = 4.00 92 = 92.01.1 = 0.100- Extra values may also be rounded and removed:-Ex: Three Sig Figs are allowed:564.71 = 565.223456 = .223- When rounding, use the number with the lowest number of Sig Figs to determine the number of Sig Figs in your answer- Ex: 110.5 (4 Sfs) x .048 (2 Sfs) = 5.304 = 5.3 (rounded with 2 Sig Figs) - When rounding, use the number with the fewest decimal places to determine the number of decimal places in your answer-Ex. 25.2 (1 decimal place) + 1.34 (2 decimal places) = 26.54 = 26.5 (rounded with one decimal place)1. % = part/whole x 100%2. Apples contain 18% (by mass) fat= 18 grams of fat/ 100 grams of appleOR= 100 grams of apple/ 18 grams of fat3. A BMI calculator indicates a person has 13% body fat. Write a Percent Factor for body fat in kg.= 13 kg fat/ 100 kg of massOR= 100 kg of mass/ 13 kg fatEx: A board has a length of 5.0 meters. What is that length in inches?- The initial unit is meters- The final unit is inchesProblem Setup and Solving:1. Write out your initial and final units2. Make a unit plan of how you will reach your final units3. Write out conversion factors and equalities which are needed in the process4. Use conversion factors to cancel out the initial unit, leaving the final unitEx: Unit 1 x Unit 2/Unit 1 = Unit 2-The two “Unit 1”(s) canceled each other out, leaving “Unit 2” as the answer- Steps to solving the ProblemIf a rock weighs 164 lb, what is its mass in kilograms?1. Initial: lb (pounds)Final: kg (kilograms)2. Lb Kg (Us Metric conversion)3. 2.20 lb = 1 KgI.20 lb/ 1 Kg OR 1kg/2.20 lb4. SOLVE:(164lb) x (1kg/2.29lb) = 74.5 kgHow many minutes are in 1.6 days?1. Initial: daysFinal: minutes2. Days  minutes 3. 1 day = 24 hours1 hour = 60 minutes(1 day/ 24 hours) OR (24 hours/ 1 day)(1 hour/ 60 Minutes) OR (60 Minutes/ 1 hour)4. SOLVE:(1.6 days) x (24 hours/1 day) x (60 minutes/ 1 hour) = 2.3 x 10^3 min- The days and hours are both canceled out, leaving the final unit minutes- Do not forget to calculate significant figures in the problem to determine how many you may have in your answer.Density of Solids through Volume Displacement:-volume is calculated from the volume of water displaced when submergedEx. To calculate the density of a rock with a mass of 4g, place it in a basin of 33mL of water. If the water rises to 45mL, what is the density of the rock?- Mass  4g- Volume 12mL (aka cm^3)- Density  Mass/Volume 4/12 1/3.33 g/cm^3- To determine if an object is more dense than water, observe whether it sinks or floats.- Water has a density of 1.00 g/mL- If an object floats, its density is less than 1.00g/mL- If an object sinks, its density is more than 1.00g/mL-How many pounds of sugar are in 233 g of candy if the candy is 50% (by mass) sugar?- Candy= 233 grams- Sugar= 50%- Find in Lb-2.20lb  1kg-1,000g  1kg- (233g candy) x (50g sugar/ 100g candy) = [116.5 g sugar]- 116.5 g sugar x (1 kg sugar/ 1,000 g sugar) = [.1165 kg sugar]- .1165 kg sugar x (2.20 lb sugar/1 kg sugar) = .2563 lb sugar-If the thickness of the skin fold at the waist indicates 9% body fat, how much fat is in a person with a mass of 140 kg?- (140 kg) x (9 kg fat/100 kg) = 1260/100 = 12.6 kg fat- Potential energy- Stored energy for use at a later time- Ex. Water behind a dam, gasoline in a gas tank- Kinetic Energy- Energy of matter in motion- Ex. Water flowing over a dam, burning gasolineCelsius to Fahrenheit:- Tc= (Tf – 32)/1.8Fahrenheit to Celsius:- Tf= 1.8 (Tc) + 32Kelvin Scale:- Tk= Tc + 273Heating Curve & Calculations:-Change in state- Heat Added-Solid (Melting point) Solid/Liquid Liquid(Boiling point) liquid/gasGasCooling Curve & Calculations:-Change in state- Heat Removed-Steam(condensation) steam/waterwater (freezing point) water/iceiceStates of Matter:-Solids- Definite shape- Definite volume- Particles are close together & vibrate slowly in fixed positions- Liquids- NO definite shape- Definite volume- Particles are close together & move slowly in random directions- Gasses- NO definite shape- NO definite volume- Take same shape and volume as their container- Fast moving particles that have little attraction to each otherEx: A thermos holds 800g of water at 55°C. If the water cools to 33°C, how many calories of heat could be transferred to someone’s body?1. Given:Mass of water: 800gInitial Temp: 55°CFinal Temp: 33°C2. NEED: Calories of heat transferred 3. Calculate Temp Change: T= Tinitial – Tfinal55°C - 33°C = 22°C4. Complete equation and solve:800g x 22°C x (1 cal/ g°C)= 1,6000 calCalorimeters:Ex.When a 7.2g sample of fat was combusted in a calorimeter, 122 kJ was absorbed. Calculate the energy value of the fat in kJ/g.(122 kJ)/(7.2g)= 16.9 kJ/gEx. (English)- Be Beryllium- Ca Calcium- H Hydrogen(Latin)- FeIron- CuCopper- AuGold- Metals blue,