Comprehensive Introductory Chemistry: Measurement, Density, Units, and Thermochemistry Vocabulary

Foundation of Chemistry and Scientific Inquiry

• Definition of Chemistry: Chemistry is defined as the branch of science that deals with the materials of the universe and the changes that these materials undergo.

• The Central Science: Chemistry is referred to as the central science because a fundamental understanding of chemical principles is essential for comprehending almost all other scientific disciplines.

• Applications and Importance of Chemistry:

  • Synthesis of novel materials.

  • Development of new pharmaceuticals.

  • Exploration and optimization of new energy sources.

  • Securing and expanding global food supplies.

  • Monitoring, protecting, and remediating the environment.

• Strategies for Learning Chemistry:

  • Master the specific scientific vocabulary.

  • Memorize foundational rules, standard units, and definitions.

  • Develop systematic problem-solving methods through continuous practice.

  • Maintain an iterative learning process by reviewing and correcting mistakes.

  • Engage in active inquiry by asking questions.

Scientific Method, Models, and Problem Solving

• Nature of Science: Science is a framework for gaining and organizing knowledge. It serves as an actionable procedure for processing, testing, and understanding natural information.

• The Scientific Approach to Problem Solving:

  1. Recognize the Problem and State It Clearly: Perform quantitative or qualitative observations of a phenomenon.

  2. Formulate a Hypothesis: Propose possible explanations or solutions based on observation.

  3. Perform Experiments: Test the proposed hypothesis or solution through controlled, reproducible procedures.

• Classification of Scientific Models:

  • Hypothesis: A tentative, testable explanation for a specific observation.

  • Theory (Model): A set of thoroughly tested hypotheses that provides an overall explanation of why natural phenomena occur. Theories attempt to explain underlying causes and evolve as new data is collected.

  • Law: A concise summary of what happens in nature, frequently expressed as a mathematical relationship. A law describes consistent natural behavior without attempting to explain the cause.

• Scientific Method Case Study (Gas Inhalation Demonstration):

  • Scenario: An instructor inhales gas from a balloon, speaks with a high-pitched voice, and asks students to identify the gas. The students observe the voice change and immediately conclude the gas is helium.

  • Analysis: The specific step of the scientific method missing from this scenario is performing experiments. The conclusion was drawn directly from an observation without testing the hypothesis.

SI Units and Measurement Systems

• Quantitative Observations (Measurements): A measurement consists of two indispensable components:

  • A number that expresses comparison or magnitude.

  • A unit that defines the scale of the measurement.

• Fundamental SI (Système International) Base Units:

  • Mass: Kilogram ($\text{kg}$)

  • Length: Meter ($\text{m}$)

  • Time: Second ($\text{s}$)

  • Temperature: Kelvin ($\text{K}$)

  • Electric Current: Ampere ($\text{A}$)

  • Amount of Substance: Mole ($\text{mol}$)

• Measurements of Length, Volume, and Mass:

  • Length: Base unit is the meter ($\text{m}$). Prefixes are attached to alter unit magnitude.

  • Volume: The measure of three-dimensional space occupied by matter. The base SI unit is the cubic meter ($\text{m}^3$). Standard laboratory volume units include the cubic centimeter ($\text{cm}^3$), liter ($\text{L}$), and milliliter ($\text{mL}$).

  • Volume Equivalences:       $1\,\text{mL} = 1\,\text{cm}^3$       $1\,\text{L} = 1\,\text{dm}^3$

  • Mass: The measure of the quantity of matter present in an object. The base SI unit is the kilogram ($\text{kg}$).

  • Mass Equivalences:       $1\,\text{kg} = 2.2046\,\text{lbs}$       $1\,\text{lb} = 453.59\,\text{g}$

• Contextual Evaluation of Common Unit Usage:

  • Reasonable Uses: A gallon of milk is approximately $4\,\text{L}$; a $200\text{-lb}$ man has a mass of approximately $90\,\text{kg}$.

  • Improper / Unreasonable Uses: Stating a basketball player has a height of $7\,\text{m}$ (excessively tall); describing a nickel as $6.5\,\text{cm}$ thick (excessively thick).

Scientific Notation and Uncertainty in Measurement

• Scientific Notation Structure: Expresses numbers as a product of a number between $1$ and $10$ multiplied by an appropriate power of $10$:   a \times 10^n \quad \text{where } 1 \le a < 10

  • Leftward Decimal Shift: Yields a positive exponent (n > 0).     $345 = 3.45 \times 10^2$

  • Rightward Decimal Shift: Yields a negative exponent (n < 0).     $0.0671 = 6.71 \times 10^{-2}$

  • Examples:     $7,882 = 7.882 \times 10^3$     $0.0000496 = 4.96 \times 10^{-5}$

• Accuracy versus Precision:

  • Accuracy: Refers to how close a measured value is to the true or theoretical value.

  • Precision: Refers to the reproducibility or agreement among a set of measurement values obtained under identical conditions.

  • Calibration Errors: An instrument that is miscalibrated can yield measurements that are highly precise (tightly clustered) but inaccurate (shifted from the true value).

• Uncertainty in Measurements:

  • Every physical measurement contains inherent uncertainty due to equipment limits or human estimation.

  • Recorded measurements must include all certain digits plus the first uncertain (estimated) digit.

  • Ruler Measurement Example: Measuring a pin whose edge falls between $2.8\,\text{cm}$ and $2.9\,\text{cm}$ yields a recorded value of $2.85\,\text{cm}$. The digits $2.8$ are certain, and the final digit $5$ is uncertain.

Rules for Significant Figures and Rounding

• Rules for Counting Significant Figures:

  1. Nonzero Integers: Always count as significant figures (e.g., $3456$ contains $4$ significant figures).

  2. Zeros:

  • Leading Zeros: Zeros that precede all non-zero digits never count as significant figures (e.g., $0.048$ contains $2$ significant figures).

  • Captive Zeros: Zeros located between non-zero digits always count as significant figures (e.g., $16.07$ contains $4$ significant figures).

  • Trailing Zeros: Zeros at the right end of a number count as significant figures only if the number explicitly contains a decimal point (e.g., $9.300$ contains $4$ significant figures; $150$ contains $2$ significant figures).

  1. Exact Numbers: Numbers obtained by counting or defined unit relationships possess an infinite number of significant figures (e.g., $1\,\text{inch} = 2.54\,\text{cm}$ exactly; $9\,\text{pencils}$).

• Exponential Notation Advantages:

  • Explicitly shows significant figure counts (e.g., $300.$ written as $3.00 \times 10^2$ indicates $3$ significant figures).

  • Reduces errors associated with recording large quantities of zeros.

• Rules for Rounding Off:

  1. If the digit to be removed is less than $5$, the preceding digit remains unchanged (e.g., $5.64 \rightarrow 5.6$).

  2. If the digit to be removed is greater than or equal to $5$, the preceding digit increases by $1$ (e.g., $5.68 \rightarrow 5.7$; $3.861 \rightarrow 3.9$).

  3. In sequential multi-step calculations, carry all calculator digits through intermediate steps and round only at the final result.

• Significant Figures in Mathematical Operations:

  • Multiplication and Division: The result contains the same number of significant figures as the measurement with the fewest significant figures.     $1.342 \times 5.5 = 7.381 \rightarrow 7.4$

  • Addition and Subtraction: The result is limited by the measurement with the fewest decimal places.

  • Graduated Cylinder Combination Example: Adding liquid from a graduated cylinder measured to the tenths place ($3.1\,\text{mL}$) to liquid from a more precise cylinder limits the overall combined volume to the tenths place due to the first cylinder.

Dimensional Analysis and Unit Conversions

• Conversion Factor Method: Multiplies the given quantity by conversion factors configured to cancel unwanted original units and retain desired units.

• Single-Unit Conversion Calculations:

  • Distance: Convert a putt of $6.8\,\text{ft}$ to inches:     $6.8\,\text{ft} \times \frac{12\,\text{in}}{1\,\text{ft}} = 81.6\,\text{in} \rightarrow 82\,\text{in}$

  • Mass: Convert an iron sample of $4.50\,\text{lbs}$ to grams ($1\,\text{kg} = 2.2046\,\text{lbs}$, $1\,\text{kg} = 1000\,\text{g}$):     $4.50\,\text{lbs} \times \frac{1\,\text{kg}}{2.2046\,\text{lbs}} \times \frac{1000\,\text{g}}{1\,\text{kg}} = 2041.19\,\text{g} \rightarrow 2040\,\text{g}$

• Applied Estimation (Driving New York to Los Angeles):

  • Required Data: Distance ($2500\,\text{miles}$), fuel efficiency ($25\,\text{miles/gallon}$), gas price (${\3.25/\text{gallon}}).

  • Calculation:     2500\,\text{miles} \times \frac{1\,\text{gallon}}{25\,\text{miles}} \times \frac{\3.25}{1\,\text{gallon}} = \$325.00$

• Combination Unit Conversions:

  • Convert $822\,\text{dm}^3/\text{s}$ to $\text{L/min}$:     $822\,\text{dm}^3/\text{s} \times \frac{1\,\text{L}}{1\,\text{dm}^3} \times \frac{60\,\text{s}}{1\,\text{min}} = 4.93 \times 10^4\,\text{L/min}$

  • Convert $0.95\,\text{kg/cm}^3$ to $\text{mg/mm}^3$:     $0.95\,\text{kg/cm}^3 \times \frac{1000\,\text{g}}{1\,\text{kg}} \times \frac{1000\,\text{mg}}{1\,\text{g}} \times \frac{1\,\text{cm}^3}{10^3\,\text{mm}^3} = 950\,\text{mg/mm}^3$

  • Convert $0.78\,\text{L/min}$ to $\text{cm}^3/\text{s}$:     $0.78\,\text{L/min} \times \frac{1000\,\text{cm}^3}{1\,\text{L}} \times \frac{1\,\text{min}}{60\,\text{s}} = 13\,\text{cm}^3/\text{s}$

Temperature Scales and Conversions

• Three Major Measuring Scales: Fahrenheit ($^\circ\text{F}$), Celsius ($^\circ\text{C}$), and Kelvin ($\text{K}$).

• Conversion Equations:   $T_K = T_C + 273.15$   $T_C = T_K - 273.15$   $T_F = 1.80(T_C) + 32$   $T_C = \frac{T_F - 32}{1.80}$

• Applied Temperature Examples:

  • Dog Body Temperature Conversion: Normal body temperature of a dog is $102^\circ\text{F}$. Convert to Kelvin:     $T_C = \frac{102 - 32}{1.80} = \frac{70.}{1.80} = 38.89^\circ\text{C}$     $T_K = 38.89 + 273.15 = 312.04\,\text{K} \rightarrow 312\,\text{K}$

  • Equivalence Point between Celsius and Fahrenheit: Find the temperature where $^\circ\text{C} = ^\circ\text{F}$.     Set $T_C = T_F = x$:     $x = 1.80x + 32$     $-0.80x = 32$     $x = -40$     Therefore, $-40^\circ\text{C} = -40^\circ\text{F}$.

Density and Mass-Volume Relationships

• Definition of Density: Density ($D$) is defined as mass ($m$) per unit volume ($V$) of a substance.   $D = \frac{m}{V}$

  • Common Units: $\text{g/cm}^3$ or $\text{g/mL}$.

• Water Displacement Method: The volume of an irregular solid is measured by submersing it in water and recording the volume displacement of liquid.

• Density Calculations:

  • Mineral Density Example: Mass $m = 17.8\,\text{g}$, volume $V = 2.35\,\text{cm}^3$.     $D = \frac{17.8\,\text{g}}{2.35\,\text{cm}^3} = 7.57\,\text{g/cm}^3$

  • Liquid Mass Example: Liquid volume $V = 49.6\,\text{mL}$, density $D = 0.85\,\text{g/mL}$.     $m = D \times V = (0.85\,\text{g/mL})(49.6\,\text{mL}) = 42.16\,\text{g} \rightarrow 42\,\text{g}$

  • Density in Specific Units Exercise: Object mass $m = 243.8\,\text{g}$, volume $V = 0.125\,\text{L}$. Calculate density in $\text{g/cm}^3$:     $V = 0.125\,\text{L} \times \frac{1000\,\text{cm}^3}{1\,\text{L}} = 125\,\text{cm}^3$     $D = \frac{243.8\,\text{g}}{125\,\text{cm}^3} = 1.9504\,\text{g/cm}^3 \rightarrow 1.95\,\text{g/cm}^3$

  • Water Displacement Level Concept Check: Copper has a density of $8.96\,\text{g/cm}^3$. A $75.0\,\text{g}$ copper sample is added to $50.0\,\text{mL}$ water in a graduated cylinder.     $V_{\text{copper}} = \frac{m}{D} = \frac{75.0\,\text{g}}{8.96\,\text{g/cm}^3} = 8.3705\,\text{cm}^3$     $\text{Final Water Level} = 50.0\,\text{mL} + 8.3705\,\text{mL} = 58.3705\,\text{mL} \rightarrow 58.4\,\text{mL}$

Energy, Temperature, and Heat

• Nature of Energy: Energy is the capacity to do work or produce heat, and is required to oppose natural forces of attraction.

• Law of Conservation of Energy: Energy can be converted from one form to another but can neither be created nor destroyed. The total energy of the universe is constant.

• Classification of Energy:

  • Potential Energy: Energy due to the position or chemical composition of an object.

  • Kinetic Energy: Energy due to the motion of an object, governed by mass and velocity.

• Thermal Definitions:

  • Temperature: A quantitative measurement of the random molecular motions of the components of a substance.

  • Heat: The flow of thermal energy between two objects driven solely by a temperature difference. Heat flows spontaneously from a hot object to a colder object.

Endothermic and Exothermic Processes

• Thermodynamic Definitions:

  • System: The primary part of the universe focused on during chemical or physical analysis.

  • Surroundings: Everything else in the universe outside the system.

• Exothermic Process:

  • Energy flows out of the system into the surroundings.

  • Energy lost by the system equals energy gained by the surroundings.

  • The potential energy of the reaction products is lower than the potential energy of the reactants.

  • Examples: Burning a match; freezing water; steam condensing on a cold pipe; hand getting cold when touching ice (system = hand).

• Endothermic Process:

  • Energy flows into the system from the surroundings.

  • Increases the potential energy of the system.

  • The potential energy of the products is higher than that of the reactants.

  • Chemical Reaction Example:     $N_2(g) + O_2(g) + \text{energy (heat)} \rightarrow 2NO(g)$

  • Physical Examples: Water boiling in a kettle; ice cream melting; ice warming when touched (system = ice).

Specific Heat Capacity and Calorimetry

• Energy Units:

  • calorie (cal): Heat required to raise the temperature of $1\,\text{g}$ of water by $1^\circ\text{C}$.

  • Joule (J): SI energy unit ($1\,\text{cal} = 4.184\,\text{J}$).

  • Calorie (Cal): Dietary unit ($1\,\text{Calorie} = 1000\,\text{calories} = 1\,\text{kcal}$).

• Energy Unit Conversion Example: Convert $60.1\,\text{cal}$ to Joules:   $60.1\,\text{cal} \times \frac{4.184\,\text{J}}{1\,\text{cal}} = 251.4584\,\text{J} \rightarrow 251\,\text{J}$

• Factors Determining Required Heat Energy:

  1. Mass of the substance being heated ($m$).

  2. Temperature change magnitude ($\Delta T = T_f - T_o$).

  3. Specific heat capacity ($C$) of the substance.

• Specific Heat Capacity ($C$): Energy required to change the temperature of $1\,\text{g}$ of a substance by $1^\circ\text{C}$.

  • Copper: $C = 0.385\,\text{J}/(\text{g}\cdot^\circ\text{C})$

  • Iron: $C = 0.451\,\text{J}/(\text{g}\cdot^\circ\text{C})$

  • Water: $C = 4.184\,\text{J}/(\text{g}\cdot^\circ\text{C})$

• Heat Formula:   $Q = m \times C \times \Delta T$

• Heat Calculation Examples:

  • Water Heating Problem: Calculate heat energy required to raise $6.25\,\text{g}$ of water from $21.0^\circ\text{C}$ to $39.0^\circ\text{C}$ ($\Delta T = 18.0^\circ\text{C}$):     $Q = (6.25\,\text{g})(4.184\,\text{J}/(\text{g}\cdot^\circ\text{C}))(18.0^\circ\text{C}) = 470.7\,\text{J}$     $\text{In calories} = 470.7\,\text{J} \times \frac{1\,\text{cal}}{4.184\,\text{J}} = 112.5\,\text{cal} \rightarrow 113\,\text{cal}$     $\text{In Calories} = 112.5\,\text{cal} \times \frac{1\,\text{Cal}}{1000\,\text{cal}} = 0.1125\,\text{Cal} \rightarrow 0.113\,\text{Cal}$

  • Iron Sample Mass Calculation: Pure iron sample requires $142\,\text{cal}$ to raise temperature from $23^\circ\text{C}$ to $92^\circ\text{C}$ ($\Delta T = 69^\circ\text{C}$, $C_{\text{iron}} = 0.45\,\text{J}/(\text{g}\cdot^\circ\text{C})$):     $Q = 142\,\text{cal} \times \frac{4.184\,\text{J}}{1\,\text{cal}} = 594.128\,\text{J}$     $m = \frac{Q}{C \times \Delta T} = \frac{594.128\,\text{J}}{(0.45\,\text{J}/(\text{g}\cdot^\circ\text{C}))(69^\circ\text{C})} = 19.13\,\text{g} \rightarrow 19\,\text{g}$

• Calorimetry and Metal Identification:

  • Enthalpy changes ($H$) are measured using a calorimeter.

  • Thermal transfer balance:     $q_{\text{lost}} = -q_{\text{gained}} \implies -(m \times C \times \Delta T)_{\text{metal}} = (m \times C \times \Delta T)_{\text{water}}$

  • Unknown Metal Problem: A $41.65\,\text{g}$ metal sample at $100.0^\circ\text{C}$ is placed in $125\,\text{g}$ of water at $25.0^\circ\text{C}$. Final equilibrium water temperature is $30.0^\circ\text{C}$.     $\Delta T_{\text{water}} = 30.0^\circ\text{C} - 25.0^\circ\text{C} = 5.0^\circ\text{C}$     $\Delta T_{\text{metal}} = 100.0^\circ\text{C} - 30.0^\circ\text{C} = 70.0^\circ\text{C}$     $C_{\text{metal}} = \frac{m_{\text{water}} \times C_{\text{water}} \times \Delta T_{\text{water}}}{m_{\text{metal}} \times \Delta T_{\text{metal}}} = \frac{(125\,\text{g})(4.184\,\text{J}/(\text{g}\cdot^\circ\text{C}))(5.0^\circ\text{C})}{(41.65\,\text{g})(70.0^\circ\text{C})} = 0.90\,\text{J}/(\text{g}\cdot^\circ\text{C})$

  • Identity: The calculated specific heat ($0.90\,\text{J}/(\text{g}\cdot^\circ\text{C})$) identifies the metal as Aluminum.

Experimental Error and Percent Error Calculations

• Percent Error Formula:   $\%\,\text{error} = \frac{|\text{Theoretical Value} - \text{Experimental Value}|}{\text{Theoretical Value}} \times 100$

• Percent Error Calculation Example:

  • Measured boiling point of water: $99.1^\circ\text{C}$

  • Accepted theoretical boiling point: $100.0^\circ\text{C}$   $\%\,\text{error} = \frac{|100.0^\circ\text{C} - 99.1^\circ\text{C}|}{100.0^\circ\text{C}} \times 100 = \frac{0.9^\circ\text{C}}{100.0^\circ\text{C}} \times 100 = 0.9\%$

• Intermediate Digit Preservation Rule:

  • When carrying a calculated value into a subsequent calculation step, retain at least two extra underlined digits to prevent cumulative round-off errors.

• Cube Density and Percent Error Sample Problem:

  • Cube mass $m = 10.932\,\text{g}$, edge length $l = 2.36\,\text{cm}$.

  • Volume $V = (2.36\,\text{cm})^3 = 13.144256\,\text{cm}^3$

  • Calculated Density: $D = \frac{10.932\,\text{g}}{13.144256\,\text{cm}^3} = 0.83169\,\text{g/cm}^3 \rightarrow 0.832\,\text{g/cm}^3$

  • Calculate Percent Error relative to accepted density $0.785\,\text{g/cm}^3$:     $\%\,\text{error} = \frac{|0.785\,\text{g/cm}^3 - 0.83169\,\text{g/cm}^3|}{0.785\,\text{g/cm}^3} \times 100 = \frac{0.04669}{0.785} \times 100 = 5.9\%$

Homework Solutions, Practice Sets, and Review Questions

• Introduction to Chemistry Review Questions:

  1. Why should a hypothesis be developed before experiments take place? To provide a structured, testable model to guide experimental design and data collection.

  2. What is the difference between a theory and a hypothesis? A hypothesis is a tentative explanation for a single observation, whereas a theory is an established model tested against multiple observations to explain why phenomena occur.

  3. What is the purpose of an experiment? To systematically test the validity of hypotheses and models.

  4. Which of the following is not part of the scientific method? A "guess" is not a valid scientific step.

  5. If experimental results disagree with an accepted theory, did you make an error? Not necessarily; experimental results may reveal that an established theory is incomplete or incorrect under specific conditions.

• Significant Figures Practice Problems:

  • Counting 1 Significant Digit: Identify measurements with exactly 1 sig fig:

  • a. $1.0 \times 10^3\,\text{m}$ ($2$ sig figs)

  • b. $2000\,\text{in}$ ($1$ sig fig)

  • c. $0.004\,\text{kg}$ ($1$ sig fig)

  • d. $\frac{|2.8\,\text{s} - 2.0\,\text{s}|}{2.0\,\text{s}} = \frac{0.8\,\text{s}}{2.0\,\text{s}} = 0.4$ ($1$ sig fig)

  • Worksheet Sig Fig Counts:

  • $502\,\text{g}$ ($3$ sig figs)

  • $0.00258\,\text{mL}$ ($3$ sig figs)

  • $0.005020\,\text{m}$ ($4$ sig figs)

  • $102,000\,\text{g}$ ($3$ sig figs)

  • $11.000\,\text{mg}$ ($5$ sig figs)

  • $100\text{ baseballs}$ (Infinite sig figs - exact counting integer)

  • Arithmetic Operations with Sig Figs:

  • $5.12\,\text{g} + 2.336\,\text{g} + 0.00258\,\text{g} = 7.45858\,\text{g} \rightarrow 7.46\,\text{g}$

  • $402.0\,\text{mL} - 10.998\,\text{mL} = 391.002\,\text{mL} \rightarrow 391.0\,\text{mL}$

  • $78.89\,\text{cm} - 40\,\text{cm} = 38.89\,\text{cm} \rightarrow 39\,\text{cm}$

  • $\frac{5.22\,\text{g}}{4.0\,\text{mL}} = 1.305\,\text{g/mL} \rightarrow 1.3\,\text{g/mL}$

  • $\frac{6.877\,\text{g}}{0.987\,\text{mL}} = 6.96758\,\text{g/mL} \rightarrow 6.97\,\text{g/mL}$

  • $(1.20\,\text{cm})(1.470\,\text{cm})(2.7\,\text{cm}) = 4.7628\,\text{cm}^3 \rightarrow 4.8\,\text{cm}^3$

• Density and Conversion Homework Set:

  • Additional Temperature Conversions:

  • Convert $-201^\circ\text{F}$ to Kelvin:       $T_C = \frac{-201 - 32}{1.80} = -129.44^\circ\text{C}$       $T_K = -129.44 + 273.15 = 143.71\,\text{K} \rightarrow 144\,\text{K}$

  • Convert $351^\circ\text{C}$ to $^\circ\text{F}$:       $T_F = 1.80(351) + 32 = 631.8 + 32 = 663.8^\circ\text{F} \rightarrow 664^\circ\text{F}$

  • Density Word Problems:

  • Mass of Wooden Block: Dimensions $4.5\,\text{cm} \times 10.2\,\text{cm} \times 2.9\,\text{cm}$, density $0.8876\,\text{g/cm}^3$       $V = (4.5)(10.2)(2.9) = 133.11\,\text{cm}^3$       $m = V \times D = (133.11\,\text{cm}^3)(0.8876\,\text{g/cm}^3) = 118.14\,\text{g} \rightarrow 120\,\text{g}$

  • Volume of Copper in Gallons: Density $8.978\,\text{g/cm}^3$, mass $22.744\,\text{kg}$ ($1.06\,\text{qt} = 1\,\text{L}$):       $m = 22744\,\text{g}$       $V = \frac{22744\,\text{g}}{8.978\,\text{g/cm}^3} = 2533.3036\,\text{cm}^3 = 2.5333\,\text{L}$       $\text{Volume in gallons} = 2.5333\,\text{L} \times \frac{1.06\,\text{qt}}{1\,\text{L}} \times \frac{1\,\text{gal}}{4\,\text{qt}} = 0.6713\,\text{gal} \rightarrow 0.671\,\text{gal}$

  • Calculated Density and Percent Error: Mass $456.2545\,\text{g}$, volume $912.509\,\text{mL}$. True density $0.511\,\text{g/mL}$.       $D = \frac{456.2545\,\text{g}}{912.509\,\text{mL}} = 0.5000000\,\text{g/mL}$       $\%\,\text{error} = \frac{|0.511 - 0.5000000|}{0.511} \times 100 = 2.15\%$

  • Mass in Pounds: Volume of $1020\,\text{cm}^3$ using calculated density $0.5000000\,\text{g/cm}^3$ ($454\,\text{g} = 1\,\text{lb}$):       $m = (1020\,\text{cm}^3)(0.5000000\,\text{g/cm}^3) = 510\,\text{g}$       $\text{Mass in lbs} = 510\,\text{g} \times \frac{1\,\text{lb}}{454\,\text{g}} = 1.123\,\text{lbs} \rightarrow 1.1\,\text{lbs}$

• Thermochemistry Practice Homework Set:

  • Unit Conversions:

  • Convert $33\,\text{mile/hr}$ to $\text{km/s}$:       $33\,\text{mi/hr} \times \frac{1.60934\,\text{km}}{1\,\text{mi}} \times \frac{1\,\text{hr}}{3600\,\text{s}} = 0.01475\,\text{km/s} \rightarrow 0.015\,\text{km/s}$

  • Convert $8.314\,\text{lbs/gal}$ to $\text{g/cm}^3$:       $8.314\,\text{lbs/gal} \times \frac{453.59\,\text{g}}{1\,\text{lb}} \times \frac{1\,\text{gal}}{3.78541\,\text{L}} \times \frac{1\,\text{L}}{1000\,\text{cm}^3} = 0.9961\,\text{g/cm}^3$

  • Heat to raise $1.54\,\text{g}$ water by $15^\circ\text{C}$:       $Q = (1.54\,\text{g})(4.184\,\text{J}/(\text{g}\cdot^\circ\text{C}))(15^\circ\text{C}) = 96.65\,\text{J} \rightarrow 97\,\text{J}$       $\text{Calories} = \frac{96.65\,\text{J}}{4.184\,\text{J/cal}} = 23.1\,\text{cal} \rightarrow 23\,\text{cal}$       $\text{Dietary Calories} = \frac{23.1\,\text{cal}}{1000\,\text{cal/Cal}} = 0.0231\,\text{Cal} \rightarrow 0.023\,\text{Cal}$

  • Convert $2.751\,\text{kJ}$ to calories:       $2751\,\text{J} \times \frac{1\,\text{cal}}{4.184\,\text{J}} = 657.50\,\text{cal} \rightarrow 657.5\,\text{cal}$

  • Convert $5.721\,\text{kcal}$ to joules:       $5.721\,\text{kcal} \times \frac{1000\,\text{cal}}{1\,\text{kcal}} \times \frac{4.184\,\text{J}}{1\,\text{cal}} = 23936.6\,\text{J} \rightarrow 23.94\,\text{kJ}$

  • Convert $12.3\,\text{nm}^3/\text{s}$ to $\text{gal/hour}$:       $12.3 \times 10^{-27}\,\text{m}^3/\text{s} \times \frac{1000\,\text{L}}{1\,\text{m}^3} \times \frac{1\,\text{gal}}{3.78541\,\text{L}} \times \frac{3600\,\text{s}}{1\,\text{hr}} = 1.17 \times 10^{-20}\,\text{gal/hr}$

  • Convert $1.25\,\text{g/dm}^3$ to $\text{mg/cm}^3$:       $1.25\,\text{g/dm}^3 \times \frac{1000\,\text{mg}}{1\,\text{g}} \times \frac{1\,\text{dm}^3}{1000\,\text{cm}^3} = 1.25\,\text{mg/cm}^3$

  • Calorimetry Mass Problem: Iron piece at $75.0^\circ\text{C}$ dropped into $50.0\,\text{g}$ water at $20.0^\circ\text{C}$. Final temperature is $28.3^\circ\text{C}$. Calculate iron mass ($C_{\text{iron}} = 0.451\,\text{J}/(\text{g}\cdot^\circ\text{C})$ ):     $q_{\text{water}} = (50.0\,\text{g})(4.184\,\text{J}/(\text{g}\cdot^\circ\text{C}))(28.3^\circ\text{C} - 20.0^\circ\text{C}) = 1736.36\,\text{J}$     $m_{\text{iron}} = \frac{1736.36\,\text{J}}{(0.451\,\text{J}/(\text{g}\cdot^\circ\text{C}))(75.0^\circ\text{C} - 28.3^\circ\text{C})} = \frac{1736.36}{(0.451)(46.7)} = 82.4\,\text{g}$

  • Bonus Calorimetry Challenge Problem: A $1.0\,\text{kg}$ ($1000\,\text{g}$) metal sample ($C = 0.50\,\text{J}/(\text{g}\cdot^\circ\text{C})$) heated to $100.0^\circ\text{C}$ is placed into $50.0\,\text{g}$ water at $20.0^\circ\text{C}$. Determine final equilibrium temperature ($T_f$):     $-(1000\,\text{g})(0.50\,\text{J}/(\text{g}\cdot^\circ\text{C}))(T_f - 100.0^\circ\text{C}) = (50.0\,\text{g})(4.184\,\text{J}/(\text{g}\cdot^\circ\text{C}))(T_f - 20.0^\circ\text{C})$     $-500(T_f - 100.0) = 209.2(T_f - 20.0)$     $-500 T_f + 50000 = 209.2 T_f - 4184$     $54184 = 709.2 T_f$     $T_f = 76.4^\circ\text{C}$