Comprehensive Study Guide for Chemistry, Measurement Conversions, Electrochemistry, and Nuclear Radiation

Academic Retention Policy Rules

• General Requirement: A student's continuation in an academic program is subject to specific academic standing requirements evaluated at the end of each semester.

• No Failing Grade: A student with no failing grade during the semester is allowed to re-enroll without restriction.

• Single Failing Grade: A student who fails in one subject at the end of a semester is allowed to re-enroll, but their permitted unit load is reduced by an amount equivalent to the number of units of the failed subject.

• Two Failing Grades & Probation:

  • A student with two failing grades at the end of the semester is placed on probation for the subsequent semester.

  • While on academic probation, the student is allowed to register for a maximum of $12\,\text{units}$.

  • Probation status is removed if the student passes all subjects taken during the probation term.

  • A student on probation who fails even one subject at the end of that semester is dropped from the rolls of the College.

• Three or More Failing Grades: A student who fails in three (3) or more subjects at the end of a semester is not allowed to re-enroll and is dropped from the rolls of the College.

• Subject Retake Limit: A student is permitted to retake the exact same subject a maximum of two times.

Unit Conversions and Dimensional Analysis

• Fundamental Concept: Measurement of physical quantities requires standardized units. When measurement units do not match desired conventions, application standards, or analytical preferences, unit conversion is necessary to interpret and apply data correctly.

  • Example: A person accustomed solely to the metric system cannot easily visualize a tree measuring $25\,\text{ft}$. Converting $25\,\text{ft}$ to meters ($7.62\,\text{m}$) provides immediate clarity.

Standard Unit Conversion Factors

Length Conversion Table

• $1\,\text{millimetre} = 0.001\,\text{metre}$

• $1\,\text{centimetre} = 0.01\,\text{metre}$

• $1\,\text{decimetre} = 0.1\,\text{metre}$

• $1\,\text{decametre} = 10\,\text{metres}$

• $1\,\text{hectometre} = 100\,\text{metres}$

• $1\,\text{kilometre} = 1000\,\text{metres}$

• $1\,\text{inch} = 2.54 \times 10^{-2}\,\text{metres}$

• $1\,\text{foot} = 0.3048\,\text{metres}$

• $1\,\text{angstrom} = 1 \times 10^{-10}\,\text{metres}$

• $1\,\text{fermi} = 1 \times 10^{-15}\,\text{metres}$

• $1\,\text{light year} = 0.946 \times 10^{16}\,\text{metres}$

• $1\,\text{mile} = 1.609344\,\text{kilometres}$

Mass Conversion Table

• $1\,\text{milligram} = 0.001\,\text{gram}$

• $1\,\text{centigram} = 0.01\,\text{gram}$

• $1\,\text{decigram} = 0.1\,\text{gram}$

• $1\,\text{decagram} = 10\,\text{grams}$

• $1\,\text{hectogram} = 100\,\text{grams}$

• $1\,\text{kilogram} = 1000\,\text{grams}$

• $1\,\text{stone} = 6350.29\,\text{grams}$

• $1\,\text{pound} = 453.592\,\text{grams}$

• $1\,\text{ounce} = 28.3495\,\text{grams}$

Volume Unit Conversions

• $1\,\text{milliliter} = 0.001\,\text{liter}$

• $1\,\text{centiliter} = 0.01\,\text{liter}$

• $1\,\text{deciliter} = 0.1\,\text{liter}$

• $1\,\text{decaliter} = 10\,\text{liters}$

• $1\,\text{hectoliter} = 100\,\text{liters}$

• $1\,\text{kiloliter} = 1000\,\text{liters}$

• $1\,\text{cubic inch} = 1.639 \times 10^{-2}\,\text{liters}$

• $1\,\text{gallon} = 3.785\,\text{liters}$

• $1\,\text{cubic foot} = 28.316\,\text{liters}$

Area Unit Conversions

• $1\,\text{sq. inch} = 6.4516 \times 10^{-4}\,\text{square metre}$

• $1\,\text{sq. foot} = 9.2903 \times 10^{-2}\,\text{square metre}$

• $1\,\text{acre} = 4.0468 \times 10^{3}\,\text{square metre}$

• $1\,\text{hectare} = 1 \times 10^{4}\,\text{square metre}$

• $1\,\text{sq. mile} = 2.5888 \times 10^{6}\,\text{square metre}$

• $1\,\text{barn} = 1 \times 10^{-28}\,\text{square metre}$

Power Unit Conversions

• $1\,\text{erg/sec} = 1 \times 10^{-5}\,\text{watt}$

• $1\,\text{BTU/hr} = 0.2930\,\text{watt}$

• $1\,\text{foot-pound/sec} = 1.356\,\text{watt}$

• $1\,\text{horsepower} = 745.7\,\text{watts}$

• $1\,\text{calorie/sec} = 4.186\,\text{watts}$

Force Unit Conversions

• $1\,\text{dyne} = 1 \times 10^{-5}\,\text{Newton}$

• $1\,\text{pound} = 4.448\,\text{Newtons}$

Density Unit Conversions

• $1\,\text{slug/cubic ft} = 515.4\,\text{kilogram/cubic metre}$

• $1\,\text{pound/cubic in} = 2.768 \times 10^{4}\,\text{kilogram/cubic metre}$

Viscosity Unit Conversions

• $1\,\text{poise} = 0.1\,\text{kg/m}\cdot\text{s}$

• $1\,\text{slug/ft} = 4.79 \times 10^{1}\,\text{kg/m}\cdot\text{s}$

Energy Unit Conversions

• $1\,\text{BTU (British thermal unit)} = 1055\,\text{Joules}$

• $1\,\text{erg} = 1 \times 10^{-7}\,\text{Joule}$

• $1\,\text{foot-pound} = 1.356\,\text{Joules}$

• $1\,\text{calorie} = 4.186\,\text{Joules}$

• $1\,\text{kilowatt-hour} = 3.6 \times 10^{6}\,\text{Joules}$

• $1\,\text{electron volt} = 1.602 \times 10^{-19}\,\text{Joule}$

• $1\,\text{liter atmosphere} = 101.13\,\text{Joules}$

Time Unit Conversions

• $1\,\text{minute} = 60\,\text{seconds}$

• $1\,\text{hour} = 60\,\text{minutes} = 3600\,\text{seconds}$

• $1\,\text{day} = 24\,\text{hours}$

• $1\,\text{week} = 7\,\text{days}$

• $1\,\text{year} = 365\,\text{days}$

Dimensional Analysis Practice Problems & Solutions

• Problem 1: Calculate the height of a person who is $5\,'3\,''$ (5 feet 3 inches) tall in meters.

  • Solution:

  • Convert feet to meters: $5\,\text{ft} \times 0.3048\,\text{m/ft} = 1.524\,\text{m}$.

  • Convert inches to meters: $3\,\text{in} \times 0.0254\,\text{m/in} = 0.0762\,\text{m}$.

  • Sum the components: $1.524\,\text{m} + 0.0762\,\text{m} = 1.6002\,\text{m}$.

• Problem 2: A student averaged $35\,\text{miles per hour}$ on a trip. What was the student's speed in feet per second?

  • Solution:

  • \n      35\,\frac{\text{mi}}{\text{hr}} \times \frac{5280\,\text{ft}}{1\,\text{mi}} \times \frac{1\,\text{hr}}{3600\,\text{s}} = \frac{184800\,\text{ft}}{3600\,\text{s}} = 51.33\,\text{ft/s}\n      

• Problem 3: If $1\,\text{pound} = 16\,\text{ounces}$, how many pounds are in $335\,\text{ounces}$?

  • Solution:

  • \n      335\,\text{oz} \times \frac{1\,\text{lb}}{16\,\text{oz}} = 20.9375\,\text{lbs}\n      

• Problem 4: Convert $9\,\text{ft}^3$ to $\text{yd}^3$, given that $1\,\text{yd} = 0.914\,\text{m}$.

  • Solution:

  • Since $1\,\text{yd} = 3\,\text{ft}$, cubing both sides yields $1\,\text{yd}^3 = 27\,\text{ft}^3$.

  • \n      9\,\text{ft}^3 \times \frac{1\,\text{yd}^3}{27\,\text{ft}^3} = 0.333\,\text{yd}^3\quad \left(\text{or } \frac{1}{3}\,\text{yd}^3\right)\n      

• Problem 5: A room measures $8\,\text{ft}$ by $14\,\text{ft}$. Calculate the area in square yards. The area is $112\,\text{ft}^2$ ($\text{Area} = \text{length} \times \text{width}$).

  • Solution:

  • Area in square feet: $8\,\text{ft} \times 14\,\text{ft} = 112\,\text{ft}^2$

  • Since $1\,\text{yd} = 3\,\text{ft}$, $1\,\text{yd}^2 = 9\,\text{ft}^2$

  • Convert to square yards: \n      112\,\text{ft}^2 \times \frac{1\,\text{yd}^2}{9\,\text{ft}^2} = 12.44\,\text{yd}^2\n      

• Problem 6: A child is prescribed a dosage of $15\,\text{mg}$ of a certain drug per day and is allowed to refill his prescription twice (total of 3 prescriptions: 1 original + 2 refills). If there are 55 tablets in a prescription, and each tablet contains $4\,\text{mg}$, how many daily doses are in the 3 prescriptions?

  • Solution:

  • Total prescriptions = $3$

  • Total tablets = $3 \times 55 = 165\,\text{tablets}$

  • Total mass of drug = $165\,\text{tablets} \times 4\,\text{mg/tablet} = 660\,\text{mg}$

  • Total daily doses = \n      \frac{660\,\text{mg}}{15\,\text{mg/day}} = 44\,\text{doses}\n      

Temperature Conversions and Scales

Temperature Formulas

• Fahrenheit to Celsius:   \n  ^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - 32)\n  

• Celsius to Fahrenheit:   \n  ^\circ\text{F} = \frac{9}{5}(^\circ\text{C}) + 32\n  

• Celsius to Kelvin:   \n  \text{K} = ^\circ\text{C} + 273.15\n  

• Fahrenheit to Kelvin:   \n  \text{K} = \frac{5}{9}(^\circ\text{F} - 32) + 273.15\n  

• Kelvin to Celsius:   \n  ^\circ\text{C} = \text{K} - 273.15\n  

• Kelvin to Fahrenheit:   \n  ^\circ\text{F} = (\text{K} - 273.15) \times \frac{9}{5} + 32\n  

Temperature Conversion Practice Problems

• Problem 1: Convert $40^\circ\text{C}$ to Kelvin.

  • Solution: \n    \text{K} = 40 + 273.15 = 313.15\,\text{K}\n    

• Problem 2: Convert $125^\circ\text{F}$ to Kelvin.

  • Solution:

  • Convert $125^\circ\text{F}$ to Celsius: \n      ^\circ\text{C} = \frac{5}{9}(125 - 32) = \frac{5}{9}(93) = 51.67^\circ\text{C}\n      

  • Convert Celsius to Kelvin: \n      \text{K} = 51.67 + 273.15 = 324.82\,\text{K}\n      

• Problem 3: Convert $275\,\text{K}$ to Celsius.

  • Solution: \n    ^\circ\text{C} = 275 - 273.15 = 1.85^\circ\text{C}\n    

Volume Formulas for Geometric Figures

Geometric Volume Formulas

• Cube:   \n  V = a^3\n     where $a = \text{length of edge or side}$.

• Rectangular Solid / Cuboid / Rectangular Prism:   \n  V = l \times w \times h\n     where $l = \text{length}$, $w = \text{width}$, $h = \text{height}$.

• Right Circular Cylinder:   \n  V = \pi r^2 h\n     where $r = \text{radius of circular base}$, $h = \text{height}$.

• General Prism:   \n  V = B \times h\n     where $B = \text{area of base}$ ($\text{side}^2$ or $l \times w$), $h = \text{height}$.

• Triangular Prism:   \n  V = \frac{b \cdot h \cdot l}{2}\n     where $b = \text{base of triangle}$, $h = \text{height of triangle}$, $l = \text{length of prism}$.

• Sphere:   \n  V = \frac{4}{3}\pi r^3\n     where $r = \text{radius of sphere}$.

• Pyramid:   \n  V = \frac{1}{3} B h\n     where $B = \text{area of base}$, $h = \text{height of pyramid}$.

• Right Circular Cone:   \n  V = \frac{1}{3}\pi r^2 h\n     where $r = \text{radius of circular base}$, $h = \text{height}$.

• Square or Rectangular Pyramid:   \n  V = \frac{1}{3} l \times w \times h\n     where $l = \text{length of base}$, $w = \text{width of base}$, $h = \text{height from base to tip}$.

• Ellipsoid:   \n  V = \frac{4}{3}\pi a b c\n     where $a, b, c = \text{semi-axes of ellipsoid}$.

• Tetrahedron:   \n  V = \frac{a^3}{6\sqrt{2}}\n     where $a = \text{length of edge}$.

Volume Unit Conversion Standards

• Imperial Standards:

  • $57.75\,\text{in}^3 = 1\,\text{qt}$

  • $4\,\text{qt} = 1\,\text{gal}$

  • $42\,\text{gal (petroleum)} = 1\,\text{barrel}$

• Metric Standards:

  • $1\,\text{cm}^3 = 1\,\text{mL}$

  • $1000\,\text{mL} = 1\,\text{liter}$

  • $1000\,\text{liters} = 1\,\text{m}^3$

• Imperial to Metric Conversions:

  • $16.39\,\text{cm}^3 = 1\,\text{in}^3$

  • $1\,\text{ft}^3 = 0.0283168\,\text{m}^3$

  • $3.79\,\text{litres} = 1\,\text{gal}$

  • $1\,\text{in}^2 = 6.45\,\text{cm}^2$

  • $1\,\text{ft}^2 = 0.092903\,\text{m}^2$

  • $1\,\text{mi}^2 = 2.59\,\text{km}^2$

Volume Practice Problems & Solutions

• Problem 1: A moving company is trying to store boxes in a storage room with a length of $8\,\text{m}$, width of $5\,\text{m}$, and height of $2.5\,\text{m}$. How many boxes can fit in this space if each box is $10\,\text{cm}$ long, $6\,\text{cm}$ wide, and $4\,\text{cm}$ high?

  • Solution:

  • Storage room volume: $V_{\text{room}} = 8\,\text{m} \times 5\,\text{m} \times 2.5\,\text{m} = 100\,\text{m}^3$

  • Convert room volume to $\text{cm}^3$: $100\,\text{m}^3 \times (100\,\text{cm/m})^3 = 100,000,000\,\text{cm}^3$

  • Box volume: $V_{\text{box}} = 10\,\text{cm} \times 6\,\text{cm} \times 4\,\text{cm} = 240\,\text{cm}^3$

  • Number of boxes = \n      \frac{100,000,000\,\text{cm}^3}{240\,\text{cm}^3/\text{box}} = 416,666.67\,\text{boxes}\n      

  • Thus, a maximum of 416,666 complete boxes can fit.

• Problem 2: Four cubes of ice with an edge of $5\,\text{cm}$ each are left to melt in a cylindrical glass with a radius of $6\,\text{cm}$. How high will the water rise when they have melted?

  • Solution:

  • Volume of one ice cube: $V_{\text{cube}} = (5\,\text{cm})^3 = 125\,\text{cm}^3$

  • Total volume of 4 ice cubes: $V_{\text{total}} = 4 \times 125\,\text{cm}^3 = 500\,\text{cm}^3$

  • Assuming melted water retains volume of $500\,\text{cm}^3$ in the glass:

  • $V_{\text{cylinder}} = \pi r^2 h \Rightarrow 500 = \pi (6)^2 h = 36\pi h$

  • Height $h = \frac{500}{36\pi} = \frac{125}{9\pi} \approx 4.42\,\text{cm}$

• Problem 3: A cube with an edge of $15\,\text{cm}$ is filled with water. Would this amount of water fit in a sphere with a $15\,\text{cm}$ radius?

  • Solution:

  • Volume of cube: $V_{\text{cube}} = (15\,\text{cm})^3 = 3375\,\text{cm}^3$

  • Volume of sphere: $V_{\text{sphere}} = \frac{4}{3}\pi (15\,\text{cm})^3 = \frac{4}{3}\pi (3375) = 4500\pi \approx 14,137.17\,\text{cm}^3$

  • Comparison: Since $3375\,\text{cm}^3 < 14,137.17\,\text{cm}^3$, yes, the water will easily fit inside the sphere.

Density, Specific Gravity, and Viscosity

Density Theory & Formulas

• Definition: Density is defined as the mass of a unit volume of a material substance. It represents the ratio between mass and volume, measuring how tightly matter is crammed together within a unit volume.

• Historical Origin: The principle of density was discovered by the ancient Greek scientist Archimedes.

• Formula: Represented by the Greek letter rho ($\rho$) or $d$:   \n  \rho = \frac{m}{V}\n     where $m = \text{mass}$ and $V = \text{volume}$.

• Units:

  • SI Unit: Kilogram per cubic meter ($\text{kg/m}^3$

  • CGS Unit: Grams per cubic centimeter ($\text{g/cm}^3$

Density Practice Problems

• Problem 1: Take a brick of salt measuring $10.0\,\text{cm} \times 10.0\,\text{cm} \times 2.0\,\text{cm}$, which weighs $433\,\text{grams}$. Calculate density.

  • Solution:

  • Volume = $10.0\,\text{cm} \times 10.0\,\text{cm} \times 2.0\,\text{cm} = 200.0\,\text{cm}^3$

  • Density = \n      \rho = \frac{433\,\text{g}}{200.0\,\text{cm}^3} = 2.165\,\text{g/cm}^3\n      

• Problem 2: You have a rock with a volume of $15\,\text{cm}^3$ and a mass of $45\,\text{g}$, and a different rock with a volume of $30\,\text{cm}^3$ and a mass of $60\,\text{g}$. Which rock is more dense? Which is less dense?

  • Solution:

  • Density of Rock 1 = $\frac{45\,\text{g}}{15\,\text{cm}^3} = 3.0\,\text{g/cm}^3$

  • Density of Rock 2 = $\frac{60\,\text{g}}{30\,\text{cm}^3} = 2.0\,\text{g/cm}^3$

  • Rock 1 is more dense ($3.0\,\text{g/cm}^3$) and Rock 2 is less dense ($2.0\,\text{g/cm}^3$).

• Problem 3: A granite boulder from the beach is $30\,\text{cm}$ on each side, giving a volume of $27,000\,\text{cm}^3$. Granite has a typical density of $2.8\,\text{g/cm}^3$. How much will this boulder weigh?

  • Solution:

  • $m = \rho \times V = 2.8\,\text{g/cm}^3 \times 27,000\,\text{cm}^3 = 75,600\,\text{g} = 75.6\,\text{kg}$

Specific Gravity (Relative Gravity)

• Definition: Specific Gravity ($SG$) is a dimensionless quantity defined as the ratio of the density of a substance to the density of a reference substance at a specified temperature.

• Reference Point for Liquids/Solids: Water at $4^\circ\text{C}$ is used as standard reference because water reaches its maximum density of $1000\,\text{kg/m}^3$ (or $1.0\,\text{g/cm}^3$) at this temperature.   \n  SG = \frac{\rho_{\text{substance}}}{\rho_{\text{water}}}\n  

• Specific Gravity of Gases: For gases, specific gravity is calculated with reference to air at specified temperature and pressure. Air density at room temperature is $1.20\,\text{kg/m}^3$.   \n  SG_{\text{gas}} = \frac{\rho_{\text{gas}}}{\rho_{\text{air}}}\n  

Specific Gravity Practice Problems

• Problem 1: If the specific gravity of ice is $0.92$, what is the density of ice?

  • Solution:

  • $\rho_{\text{ice}} = SG \times \rho_{\text{water}} = 0.92 \times 1000\,\text{kg/m}^3 = 920\,\text{kg/m}^3$ (or $0.92\,\text{g/cm}^3$).

• Problem 2: If the density of gold is $19,300\,\text{kg/m}^3$, what is the specific gravity of gold?

  • Solution:

  • $SG = \frac{19,300\,\text{kg/m}^3}{1000\,\text{kg/m}^3} = 19.3$

• Problem 3: You have a sample of granite with density $2.8\,\text{g/cm}^3$. The density of water is $1.0\,\text{g/cm}^3$. What is the specific gravity of your granite?

  • Solution:

  • $SG = \frac{2.8\,\text{g/cm}^3}{1.0\,\text{g/cm}^3} = 2.8$

• Problem 4: You have a sample of granite with density $174.8\,\text{lbs/ft}^3$. The density of water is $62.4\,\text{lbs/ft}^3$. What is the specific gravity of the granite?

  • Solution:

  • $SG = \frac{174.8\,\text{lbs/ft}^3}{62.4\,\text{lbs/ft}^3} = 2.8$

Viscosity

• Definition: The viscosity of a fluid is a measure of its resistance to deformation or flow at a given rate.

Specific Heat Capacity and Thermal Physics

Core Definitions

• Heat: Energy transferred from one body to another as the result of a difference in temperature.

• Specific Heat: The quantity of heat required to raise the temperature of one gram of a substance by one Celsius degree ($1^\circ\text{C}$).

  • Units: Calories per gram per Celsius degree ($\text{cal/g}\cdot^\circ\text{C}$) or Joules per gram per Celsius degree ($\text{J/g}\cdot^\circ\text{C}$).

  • Water Standard: Specific heat of water is $1\,\text{calorie/g}\cdot^\circ\text{C}$ or $4.186\,\text{Joules/g}\cdot^\circ\text{C}$ ($4200\,\text{J/kg}\cdot^\circ\text{C}$).

Specific Heat Formula

\n\Delta Q = m c \Delta T\n where:

• $\Delta Q = \text{heat gained or lost (in Joules)}$

• $m = \text{mass (in kg or g)}$

• $c = \text{specific heat capacity}$

• $\Delta T = \text{temperature difference} = T_f - T_i$ (where $T_f$ is final temperature and $T_i$ is initial temperature in $^\circ\text{C}$ or $\text{K}$)

Key Considerations & Rules

1. Ensure the change in temperature $\Delta T$ is evaluated as a positive quantity.

2. Convert all problem variables to consistent matching unit sets before calculation.

3. If specific heat is provided in grams ($c_p = 0.41868\,\text{J/g}\cdot^\circ\text{C}$), either convert mass to grams or convert $c_p$ to kilograms ($418.68\,\text{J/kg}\cdot^\circ\text{C}$).

4. If given $c_p$ in $\text{kJ}$, unit conversion to $\text{J}$ yields Joules; otherwise leaving $\text{kJ}$ yields answers directly in $\text{kJ}$.

5. Heat Capacity vs. Specific Heat Capacity: Specific heat capacity is mass-normalized (per unit mass). Heat Capacity ($C_p$) represents the total system heat requirement and is calculated as:    \n   C_p = m \times c\n   

6. The heat capacity of a given substance is treated as constant across standard working temperature differences.

Specific Heat Practice Problems & Solutions

• Problem 1: Calculate the heat required to raise $0.6\,\text{kg}$ of sand from $30^\circ\text{C}$ to $90^\circ\text{C}$. (Specific heat of sand = $830\,\text{J/kg}\cdot^\circ\text{C}$).

  • Solution:

  • $\Delta T = 90 - 30 = 60^\circ\text{C}$

  • $\Delta Q = m c \Delta T = 0.6\,\text{kg} \times 830\,\text{J/kg}\cdot^\circ\text{C} \times 60^\circ\text{C} = 29,880\,\text{Joules}$

• Problem 2: Compute the temperature difference if $50\,\text{kg}$ of water absorbs $500\,\text{kJ}$ of heat.

  • Solution:

  • $\Delta Q = 500\,\text{kJ} = 500,000\,\text{Joules}$

  • $c_{\text{water}} = 4186\,\text{J/kg}\cdot^\circ\text{C}$

  • $\Delta T = \frac{\Delta Q}{m c} = \frac{500,000}{50 \times 4186} = \frac{500,000}{209,300} \approx 2.39^\circ\text{C}$

• Problem 3: Calculate the specific heat of water if $4200\,\text{J}$ is needed to change the temperature of $1\,\text{kg}$ by $1^\circ\text{C}$.

  • Solution:

  • $c = \frac{\Delta Q}{m \Delta T} = \frac{4200\,\text{J}}{1\,\text{kg} \times 1^\circ\text{C}} = 4200\,\text{J/kg}\cdot^\circ\text{C}$

• Problem 4: How much heat is necessary to heat up $0.05\,\text{kg}$ of Cobalt from $30^\circ\text{C}$ to $40^\circ\text{C}$? The specific heat of Cobalt is $0.41868\,\text{J/g}\cdot^\circ\text{C}$.

  • Solution:

  • $m = 0.05\,\text{kg} = 50\,\text{g}$

  • $\Delta T = 40 - 30 = 10^\circ\text{C}$

  • $\Delta Q = 50\,\text{g} \times 0.41868\,\text{J/g}\cdot^\circ\text{C} \times 10^\circ\text{C} = 209.34\,\text{Joules}$

Electrochemical Energy and Oxidation States

• Electrochemical Energy Definition: Electrochemical energy is defined as the energy converting electrical energy to chemical energy and vice versa. It governs fuel cells, photoelectrochemical systems, and energy storage devices such as batteries, supercapacitors, and ultra-capacitors.

Oxidation Numbers

• Definition: An oxidation number is a positive, negative, or zero integer assigned to an atom in a chemical species. It keeps track of the electron density gained or lost by an atom when combining to form chemical compounds, enabling the balancing of complex redox reactions.

8 Fundamental Rules for Assigning Oxidation Numbers

1. Pure Elemental Form: In its free/pure elemental state (uncombined with other elements), an atom's oxidation number is always 0.

   • Examples: $\text{Cl}_2^0$, $\text{Ag}^0$, $\text{O}_2^0$, $\text{Fe}(s)$.

2. Monatomic Ions: The oxidation number of a monatomic ion equals its net ionic charge.

   • Examples: $\text{K}^+ = +1$, $\text{Br}^- = -1$, $\text{Fe}^{2+} = +2$.

3. Group 1 and Group 2 Metals: Metals in Group 1 have an oxidation state of +1; Group 2 metals have an oxidation state of +2 in compounds. (Hydrogen is excluded as it is a nonmetal).

4. Hydrogen:

   • Oxidation state is +1 when bonded to nonmetals (e.g., $\text{HCl}$, $\text{H}_2\text{O}$) because hydrogen is less electronegative than most nonmetals.

   • Oxidation state is -1 when bonded to metals in metal hydrides (e.g., $\text{NaH}$, $\text{CaH}_2$) because hydrogen is more electronegative than metals.

5. Oxygen:

   • Generally has an oxidation state of -2 in oxide compounds ($\text{O}^{2-}$).

   • Exception 1: In peroxides ($\text{O}_2^{2-}$, e.g., $\text{H}_2\text{O}_2$), oxygen has an oxidation state of -1.

   • Exception 2: In superoxides ($\text{O}_2^-$), oxygen has an oxidation state of -1/2.

   • Exception 3: In oxygen difluoride ($\text{OF}_2$), fluorine is more electronegative, giving oxygen an oxidation state of +2.

6. Fluorine: Fluorine is the most electronegative element ($\text{Pauling value} = 4.0$) and always carries an oxidation number of -1 in all compounds ($\text{HF}$, $\text{SF}_6$, $\text{NaF}$, $\text{CaF}_2$).

7. Other Halogens (Cl, Br, I): Halogens usually carry an oxidation number of -1, except when combined with oxygen or fluorine where they exhibit positive oxidation states (e.g., in $\text{HClO}$, $\text{Cl} = +1$; in $\text{HClO}_4$, $\text{Cl} = +7$).

8. Sum of Oxidation States:

   • Neutral compound sum = 0.

   • Polyatomic ion sum = net charge of the ion.

Electronegativity Trends

• Definition: Electronegativity is the measure of the ability of an atom in a chemical bond to attract shared electrons to itself. The most electronegative element determines which atom carries the negative formal charge.

• Pauling Scale Values:

  • Fluorine ($\text{F}$) = $4.0$

  • Oxygen ($\text{O}$) = $3.5$

  • Nitrogen ($\text{N}$) = $3.0$

  • Chlorine ($\text{Cl}$) = $3.0$

  • Bromine ($\text{Br}$) = $2.8$

  • Carbon ($\text{C}$) = $2.5$

  • Sulfur ($\text{S}$) = $2.5$

  • Iodine ($\text{I}$) = $2.5$

  • Hydrogen ($\text{H}$) = $2.1$

  • Phosphorus ($\text{P}$) = $2.1$

  • Boron ($\text{B}$) = $2.0$

Oxidation State Determination Practice Problems

• Problem 1: Find the oxidation number of Sulfur in $\text{H}_2\text{SO}_4$

  • Solution: $2(+1) + x + 4(-2) = 0 \Rightarrow 2 + x - 8 = 0 \Rightarrow x = +6$

• Problem 2: Find the oxidation number of Chromium in $\text{Cr}_2\text{O}_7^{2-}$

  • Solution: $2x + 7(-2) = -2 \Rightarrow 2x - 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6$

• Problem 3: Find the oxidation number of Manganese in $\text{MnO}_4^{2-}$

  • Solution: $x + 4(-2) = -2 \Rightarrow x - 8 = -2 \Rightarrow x = +6$

• Problem 4: Find the oxidation number of Chromium in $\text{Cr}_2\text{O}_3$

  • Solution: $2x + 3(-2) = 0 \Rightarrow 2x = +6 \Rightarrow x = +3$

  • Observation: Transition elements exhibit variable oxidation states depending on compound composition ($+6$ in $\text{Cr}_2\text{O}_7^{2-}$ vs $+3$ in $\text{Cr}_2\text{O}_3$).

• Problem 5: Determine oxidation states in $\text{Fe}(s) + \text{O}_2(g) \rightarrow \text{Fe}_2\text{O}_3(g)$

  • Solution: $\text{Fe}(s) = 0$, $\text{O}_2(g) = 0$. In $\text{Fe}_2\text{O}_3$, $\text{O} = -2$, so $2x + 3(-2) = 0 \Rightarrow x = +3$ ($\text{Fe} = +3$).

• Problem 6: Determine oxidation states in $\text{Ag}(s) + \text{H}_2\text{S} \rightarrow \text{Ag}_2\text{S}(g) + \text{H}_2(g)$

  • Solution: $\text{Ag}(s) = 0$; in $\text{H}_2\text{S}$, $\text{H} = +1$, $\text{S} = -2$; in $\text{H}_2(g)$, $\text{H} = 0$; in $\text{Ag}_2\text{S}$, $\text{S} = -2$, so $2x - 2 = 0 \Rightarrow x = +1$ ($\text{Ag} = +1$).

• Problem 7: Find oxidation state of Phosphorus in $\text{Na}_3\text{PO}_3$

  • Solution: $3(+1) + x + 3(-2) = 0 \Rightarrow 3 + x - 6 = 0 \Rightarrow x = +3$

• Problem 8: Find oxidation state of Phosphorus in $\text{H}_2\text{PO}_4^-$

  • Solution: $2(+1) + y + 4(-2) = -1 \Rightarrow 2 + y - 8 = -1 \Rightarrow y = +5$

Oxidation-Reduction (Redox) Reactions and Agents

Redox Concepts

• Definition: An oxidation-reduction (redox) reaction is a chemical reaction involving the transfer of electrons between chemical species, causing changes in oxidation numbers.

  • Vital life and industrial processes driven by redox include photosynthesis, cellular respiration, combustion, and metallic corrosion.

• OIL RIG Mnemonic:

  • Oxidation Is Loss of electrons (Oxidation number increases).

  • Reduction Is Gain of electrons (Oxidation number decreases).

Agents Defined

• Oxidizing Agent (Oxidant): The substance that gains electrons and undergoes reduction. It causes the oxidation of another species, decreasing its own oxidation number.

• Reducing Agent (Reductant): The substance that loses electrons and undergoes oxidation. It causes the reduction of another species, increasing its own oxidation number.

Agent Identification Step-by-Step Procedure

1. Assign oxidation numbers to every element on reactant and product sides.

2. Identify species whose oxidation numbers change.

3. Determine oxidation (number goes up) and reduction (number goes down).

4. Assign agents: Oxidized species = Reducing Agent; Reduced species = Oxidizing Agent.

Worked Example 1: $\text{Ca} + \text{Cl}_2 \rightarrow \text{CaCl}_2$

• Oxidation numbers: $\text{Ca}^0 + \text{Cl}_2^0 \rightarrow \text{Ca}^{2+}\text{Cl}_2^{-1}$

• Calcium: $0 \rightarrow +2$ (Increases $\rightarrow$ Oxidation $\rightarrow$ Reducing Agent)

• Chlorine: $0 \rightarrow -1$ (Decreases $\rightarrow$ Reduction $\rightarrow$ Oxidizing Agent)

Worked Example 2: $\text{Cl}_2(aq) + 2\text{Br}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{Br}_2(aq)$

• Oxidation half-reaction: $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$ ($-1 \rightarrow 0$, lost electrons $\rightarrow$ $\text{Br}^-$ is the Reducing Agent).

• Reduction half-reaction: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ ($0 \rightarrow -1$, gained electrons $\rightarrow$ $\text{Cl}_2$ is the Oxidizing Agent).

Worked Example 3: $\text{MnO}_4^- + \text{SO}_3^{2-} \rightarrow \text{Mn}^{2+} + \text{SO}_4^{2-}$

• Sulfur in $\text{SO}_3^{2-}$ ($+4$) to $\text{SO}_4^{2-}$ ($+6$): Oxidized $\rightarrow$ $\text{SO}_3^{2-}$ is the Reducing Agent.

• Manganese in $\text{MnO}_4^-$ ($+7$) to $\text{Mn}^{2+}$ ($+2$): Reduced $\rightarrow$ $\text{MnO}_4^-$ is the Oxidizing Agent.

Balancing Chemical and Redox Equations

Law of Conservation of Mass

• Fundamental Principle: Chemical equations must be balanced to satisfy the Law of Conservation of Mass/Matter, which states that in a closed system, matter is neither created nor destroyed.

• Rules for Equation Manipulation: Coefficients (numbers in front of formulas) are changed to adjust atom counts. Subscripts within chemical formulas must NEVER be altered, as doing so changes the chemical identity of the substance ($\text{H}_2\text{O} \neq \text{H}_2\text{O}_2$).

Combustion Balancing Example ($\text{CH}_4$)

• Unbalanced: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$

  • Reactants: $1\,\text{C}$, $4\,\text{H}$, $2\,\text{O}$

  • Products: $1\,\text{C}$, $2\,\text{H}$, $3\,\text{O}$ (Unbalanced)

• Balanced: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

  • Reactants & Products: $1\,\text{C}$, $4\,\text{H}$, $4\,\text{O}$ (Satisfies Conservation of Mass).

  • Molar Interpretation: 1 mole of methane plus 2 moles of oxygen produce 1 mole of carbon dioxide and 2 moles of water.

Half-Reaction Method for Balancing Redox Reactions in Acidic Medium

1. Write Half-Reactions: Separate the skeleton equation into oxidation and reduction half-reactions.

2. Balance Key Atoms: Balance all atoms other than Oxygen and Hydrogen.

3. Balance Oxygen: Add $\text{H}_2\text{O}$ molecules to the side deficient in oxygen.

4. Balance Hydrogen: Add $\text{H}^+$ ions to the side deficient in hydrogen.

5. Balance Charge: Add electrons ($e^-$) to the side with greater positive net charge.

6. Equalize Electrons & Combine: Multiply each half-reaction by integers so electron loss equals electron gain, add the half-reactions, and cancel identical terms on both sides.

Comprehensive Worked Step-by-Step Example

Balance: $\text{Cr}_2\text{O}_7^{2-}(aq) + \text{Cl}^-(aq) \rightarrow \text{Cr}^{3+}(aq) + \text{Cl}_2(aq)$

• Step 1: Write half-reactions

  • Reduction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$

  • Oxidation: $\text{Cl}^- \rightarrow \text{Cl}_2$

• Step 2: Balance non-O and non-H atoms

  • $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$

  • $2\text{Cl}^- \rightarrow \text{Cl}_2$

• Step 3: Balance Oxygens with $\text{H}_2\text{O}$

  • $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

• Step 4: Balance Hydrogens with $\text{H}^+$

  • $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

• Step 5: Balance Charge with electrons ($e^-$)

  • Reduction side net charge: $14(+1) - 2 = +12$. Product charge: $2(+3) = +6$. Add $6e^-$ to reactants:     $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

  • Oxidation side net charge: $2(-1) = -2$. Product charge: $0$. Add $2e^-$ to products:     $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$

• Step 6: Equalize electrons and sum half-reactions

  • Multiply oxidation half-reaction by 3:     $3(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-) \Rightarrow 6\text{Cl}^- \rightarrow 3\text{Cl}_2 + 6e^-$

  • Sum half-reactions:     \n    \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Cl}^-(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cl}_2(aq) + 7\text{H}_2\text{O}(l)\n    

Electrochemistry: Cells, Batteries, Corrosion, and Electrolysis

Electrochemical Cells & Batteries

• Electrochemistry Definition: The scientific discipline dealing with the relationship between electrical energy and chemical transformations.

• Battery Definition: A device consisting of one or several electrochemical cells connected in series/parallel that stores chemical energy and converts it into electricity via spontaneous redox reactions.

Galvanic (Voltaic) Cell Mechanism

• Named After: Luigi Galvani and Alessandro Volta.

• Operation: Converts chemical energy into electrical energy via spontaneous redox reactions.

• Component Roles in a Zinc-Copper Galvanic Cell:

  1. Anode (Zinc Strip): Site of oxidation. Zinc loses electrons:      $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$

  2. External Circuit: Electrons travel from the zinc anode through the external wire toward the copper cathode, generating electric current.

  3. Cathode (Copper Strip): Site of reduction. Aqueous copper ions gain incoming electrons:      $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

  4. Salt Bridge: Contains aqueous electrolyte solution allowing migration of positive cations and negative anions to maintain charge neutrality in both half-cell compartments.

Corrosion vs. Explosions

• Corrosion: Gradual deterioration of metals resulting from chemical or electrochemical reactions with their environment (e.g., rusting of iron via redox).

• Explosions: Extremely rapid chemical reactions accompanied by instantaneous release of high energy in the form of extreme heat, light, sound, and rapidly expanding gases.

Oxygen as an Oxidizing Agent

• Properties: Highly electronegative element, stable in elemental diatomic form ($\text{O}_2$).

• Abundance:

  • Atmosphere: 2nd most abundant atmospheric gas ($20.95\%$

  • Earth's Crust: Most abundant element ($46.4\%$

  • Human Body: Most abundant element by mass ($65.0\%$

Alkali Metals as Reducing Agents

• Group 1 Elements: Hydrogen ($\text{H}$), Lithium ($\text{Li}$), Sodium ($\text{Na}$), Potassium ($\text{K}$), Rubidium ($\text{Rb}$), Cesium ($\text{Cs}$), Francium ($\text{Fr}$).

• Reducing Strength: Alkali metals favor losing their single valence electron to obtain a noble gas electron configuration, making them exceptionally powerful reducing agents.

Standard Reduction Potentials ($E^\circ$

• $\text{Li}^+ + e^- \rightarrow \text{Li}\quad (-3.04\,\text{V})$ [Strongest Reducing Agent]

• $\text{Na}^+ + e^- \rightarrow \text{Na}\quad (-2.71\,\text{V})$

• $\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}\quad (-2.38\,\text{V})$

• $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\quad (-1.66\,\text{V})$

• $2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-\quad (-0.83\,\text{V})$

• $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}\quad (-0.74\,\text{V})$

• $\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}\quad (-0.41\,\text{V})$

• $2\text{H}^+ + 2e^- \rightarrow \text{H}_2\quad (0.00\,\text{V})$ [Standard Reference]

• $\text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+}\quad (+0.15\,\text{V})$

• $\text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+\quad (+0.16\,\text{V})$

• $\text{Ag}^+ + e^- \rightarrow \text{Ag}\quad (+0.80\,\text{V})$

• $\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-\quad (+1.07\,\text{V})$

• $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\quad (+1.36\,\text{V})$

• $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\quad (+1.49\,\text{V})$ [Strong Oxidizing Agent]

Electrolysis

• Definition: The non-spontaneous process of driving a chemical decomposition reaction by passing a direct electric current through an ionic compound in solution or molten state.

• Electrode Reactions:

  • Anode (Positive Electrode): Attracts negatively charged anions; site of oxidation.

  • Cathode (Negative Electrode): Attracts positively charged cations; site of reduction.

• Electrochemical Water Splitting: Passing current through water decomposes it into oxygen gas at the anode and hydrogen gas at the cathode.

• Industrial Applications:

  1. Extraction of reactive metals from ores

  2. Electro-refining of crude metals

  3. Manufacturing of chemical reagents

  4. Electroplating

  5. Electro-decomposition of rubber

  6. Electro-metallization & Electro-facing

Nuclear Chemistry and Radiation Equations

Laws of Nuclear Balancing

• Nuclear equations represent rearrangements of subatomic particles (protons and neutrons) rather than intact atoms. They are balanced according to two strict conservation rules:

  1. Conservation of Mass Number: The sum of mass numbers ($A$) of reactants equals the sum of mass numbers of products.

  2. Conservation of Atomic Number / Charge: The sum of atomic numbers/charges ($Z$) of reactants equals the sum of atomic numbers/charges of products.

Nuclear Notation Standard

\n^{A}_{Z}\text{X}\n where:

• $\text{X} = \text{element chemical symbol}$

• $A = \text{total number of nucleons (protons } + \text{ neutrons)} = \text{mass number}$

• $Z = \text{total number of protons} = \text{atomic number}$

5 Types of Radioactive Decay

1. Alpha Decay ($\alpha$)

• Process: An unstable heavy parent nucleus ejects an alpha particle ($^{4}_{2}\text{He}$), containing 2 protons and 2 neutrons.

• Effect: Mass number $A$ decreases by 4; Atomic number $Z$ decreases by 2.

• General Equation: \n  ^{A}_{Z}\text{X} \rightarrow ^{A-4}_{Z-2}\text{Y} + ^{4}_{2}\text{He}\n  

• Primary Examples:

  • Decay of Uranium-238: \n    ^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}\n    

  • Decay of Neptunium-237: \n    ^{237}_{93}\text{Np} \rightarrow ^{233}_{91}\text{Pa} + ^{4}_{2}\text{He}\n    

  • Decay of Platinum-175: \n    ^{175}_{78}\text{Pt} \rightarrow ^{171}_{76}\text{Os} + ^{4}_{2}\text{He}\n    

• Alpha Decay Practice Problems & Solutions:

  1. $^{239}<em>{94}\text{Pu} \rightarrow ^{4}</em>{2}\text{He} + \mathbf{^{235}_{92}\text{U}}$

  2. $^{228}<em>{88}\text{Ra} \rightarrow ^{4}</em>{2}\text{He} + \mathbf{^{224}_{86}\text{Rn}}$

  3. $^{241}<em>{95}\text{Am} \rightarrow ^{4}</em>{2}\text{He} + \mathbf{^{237}_{93}\text{Np}}$

  4. $^{213}<em>{83}\text{Bi} \rightarrow ^{4}</em>{2}\alpha + \mathbf{^{209}_{81}\text{Tl}}$

  5. $^{195}<em>{78}\text{Pt} \rightarrow ^{4}</em>{2}\alpha + \mathbf{^{191}_{76}\text{Os}}$

2. Beta Decay ($\beta^-$)

• Process: A neutron in the nucleus converts into a proton, emitting a high-energy electron ($_{-1}^{0}e$ or $\beta^-$) and an electron antineutrino ($\bar{\nu}_e$).

• Effect: Mass number $A$ is unchanged; Atomic number $Z$ increases by 1.

• General Equation: \n  ^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y} + _{-1}^{0}e + \bar{\nu}_e\n  

• Example (Carbon-14): \n  ^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + _{-1}^{0}e + \bar{\nu}_e\n  

• Beta Decay Practice Problems & Solutions:

  1. $^{85}<em>{36}\text{Kr} \rightarrow \mathbf{^{85}</em>{37}\text{Rb}} + _{-1}^{0}e$

  2. $^{87}<em>{37}\text{Rb} \rightarrow \mathbf{^{87}</em>{38}\text{Sr}} + _{-1}^{0}e$

  3. $^{32}<em>{15}\text{P} \rightarrow \mathbf{^{32}</em>{16}\text{S}} + _{-1}^{0}e$

  4. $^{216}<em>{84}\text{Po} \rightarrow \mathbf{^{216}</em>{85}\text{At}} + _{-1}^{0}e$

  5. $^{210}<em>{83}\text{Bi} \rightarrow \mathbf{^{210}</em>{84}\text{Po}} + _{-1}^{0}e$

3. Gamma Decay ($\gamma$)

• Process: A nucleus in an excited energy state releases excess binding energy as a high-energy gamma photon ($^{0}_{0}\gamma$) to transition into its ground energy state. No conversion of subatomic particles occurs.

• Effect: Mass number $A$ is unchanged; Atomic number $Z$ is unchanged.

• General Equation: \n  ^{A}_{Z}\text{X}^* \rightarrow ^{A}_{Z}\text{X} + ^{0}_{0}\gamma\n  

• Gamma Decay Practice Problems & Solutions:

  1. $^{222}<em>{86}\text{Rn} \rightarrow \mathbf{^{222}</em>{86}\text{Rn}} + ^{0}_{0}\gamma$

  2. $^{99}<em>{43}\text{Tc} \rightarrow \mathbf{^{99}</em>{43}\text{Tc}} + ^{0}_{0}\gamma$

  3. $^{60}<em>{27}\text{Co} \rightarrow \mathbf{^{60}</em>{27}\text{Co}} + ^{0}_{0}\gamma$

4. Positron Emission ($\beta^+$ Decay)

• Process: A nuclear proton disintegrates into a neutron and emits a positron ($_{+1}^{0}e$, a positively charged electron).

• Effect: Mass number $A$ is unchanged; Atomic number $Z$ decreases by 1.

• General Equation: \n  ^{A}_{Z}\text{X} \rightarrow ^{A}_{Z-1}\text{Y} + _{+1}^{0}e\n  

5. Electron Capture

• Process: An unstable nucleus captures an inner-shell orbital electron ($_{-1}^{0}e$), combining it with a proton to form a neutron and releasing an X-ray photon.

• Effect: Mass number $A$ is unchanged; Atomic number $Z$ decreases by 1.

• General Equation: \n  ^{A}_{Z}\text{X} + _{-1}^{0}e \rightarrow ^{A}_{Z-1}\text{Y}\n  

• Electron Capture Practice Problems & Solutions:

  1. $^{40}<em>{19}\text{K} + </em>{-1}^{0}e \rightarrow \mathbf{^{40}_{18}\text{Ar}}$

  2. $^{37}<em>{18}\text{Ar} + </em>{-1}^{0}e \rightarrow \mathbf{^{37}_{17}\text{Cl}}$

  3. $^{86}<em>{40}\text{Zr} + </em>{-1}^{0}e \rightarrow \mathbf{^{86}_{39}\text{Y}}$

  4. $^{204}<em>{84}\text{Po} + </em>{-1}^{0}e \rightarrow \mathbf{^{204}_{83}\text{Bi}}$

  5. $^{57}<em>{27}\text{Co} + </em>{-1}^{0}e \rightarrow \mathbf{^{57}_{26}\text{Fe}}$