Grade 8 Intermediate Science Trimester 3 Assessment Review Notes

Chemical Reaction Types and Balancing Equations

• General Types of Chemical Reactions:     * Synthesis (Combination): A reaction where two or more substances combine to form a single new substance ($A + B \rightarrow AB$).         * Example: $CaO + H_2O \rightarrow Ca(OH)_2$. In this diagram, substance A and substance B combine to form a single product.         * Example: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$. This is a synthesis reaction where magnesium and oxygen combine.     * Decomposition: A single compound breaks down into two or more simpler substances ($AB \rightarrow A + B$).     * Single Replacement: One element replaces another element in a compound ($A + BC \rightarrow AC + B$).         * Example: $Mg(s) + 2AgNO_3(aq) \rightarrow Mg(NO_3)_2(aq) + 2Ag(s)$.     * Double Replacement: The ions of two compounds exchange places in an aqueous solution to form two new compounds ($AB + CD \rightarrow AD + CB$).

• Balancing Chemical Equations:     * The Law of Conservation of Mass must be satisfied, meaning the number of atoms for each element on the reactant side must equal the number of atoms on the product side.     * Case Study: Iron(III) Oxide and Carbon:         * Unbalanced equation: $Fe_2O_3 + C \rightarrow Fe + CO_2$         * Balancing process:             * Reactants: $Fe = 2$, $O = 3$, $C = 1$             * Products: $Fe = 1$, $O = 2$, $C = 1$         * To balance oxygen, use a coefficient of 2 for $Fe_2O_3$ and 3 for $CO_2$ (resulting in 6 oxygen atoms on both sides).         * Then balance carbon: 3 carbon atoms on the product side means a coefficient of 3 for $C$.         * Finally balance iron: 4 iron atoms on the reactant side ($2 \times 2$) means a coefficient of 4 for $Fe$.         * Balanced equation: $2Fe_2O_3 + 3C \rightarrow 4Fe + 3CO_2$         * The correct coefficient for Fe is 4.

Stoichiometry and Limiting Reactants

• Mole-to-Mole Ratios:     * Reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$     * Ratio of $H_2O_2$ to $O_2$ is $2:1$.     * To produce $3\,\text{moles}$ of $O_2$, the calculation is $3 \times 2 = 6\,\text{moles}$ of $H_2O_2$ required.

• Mass-to-Mole-to-Mass Conversions:     * Example: Potassium and Hydrogen Gas:         * Reaction: $2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)$         * Goal: Produce $0.50\,\text{mol}$ of $H_2$ gas.         * Ratio of $H_2$ to $K$ is $1:2$.         * Moles of K needed = $0.50\,\text{mol} \times 2 = 1.0\,\text{mol}$.         * Atomic mass of K = $39\,\text{g/mol}$.         * Mass of K = $1.0\,\text{mol} \times 39\,\text{g/mol} = 39\,\text{g}$.

• Limiting Reactant Analysis:     * Example: Mg and HCl:         * Reaction: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$         * Given: $20\,\text{mL}$ of $1.0\,\text{M HCl}$ and $0.24\,\text{g}$ of powdered Mg.         * Moles of HCl: $0.020\,\text{L} \times 1.0\,\text{mol/L} = 0.02\,\text{mol}$.         * Moles of Mg: $\frac{0.24\,\text{g}}{24\,\text{g/mol}} = 0.01\,\text{mol}$.         * The required ratio of $HCl:Mg$ is $2:1$.         * Since $0.02\,\text{mol}$ of HCl reacts exactly with $0.01\,\text{mol}$ of Mg, neither is in excess, and the reaction is stoichiometric.

Intermolecular Forces (IMF) and Properties of Matter

• London Dispersion Forces (LDF):     * The weakest IMF, present in all molecules, especially nonpolar ones.     * Strength increases with molecular size (number of electrons) and surface area.     * Noble Gases comparison: Boiling points increase from Neon ($-246.1^{\circ}\text{C}$) to Argon ($-185.7^{\circ}\text{C}$) to Krypton ($-153.2^{\circ}\text{C}$) because larger atoms have stronger LDF.     * Gas comparison: LDF are weakest for $H_2$ compared to $O_2$, $Cl_2$, or $F_2$ because $H_2$ has the smallest surface area and the fewest electrons.

• Hydrogen Bonding:     * A strong IMF occurring when hydrogen is bonded to N, O, or F.     * Water (H₂O): Exhibits strong hydrogen bonding, leading to high surface tension and high boiling points ($100^{\circ}\text{C}$).     * Ammonia (NH₃): Formed between the partial negative charge of Nitrogen and the partial positive charge of Hydrogen on an adjacent molecule.

• Dipole-Dipole Interactions:     * Occur between polar molecules (e.g., $PCl_3$, $HCl$, $CH_3Cl$).     * These are stronger than LDF but generally weaker than hydrogen bonds.

• Ion-Dipole Attractions:     * Occur when ionic compounds (like NaCl) dissolve in water.     * Partially negative oxygen ends of water molecules surround positive cations ($Na^+$).     * Partially positive hydrogen ends of water molecules surround negative anions ($Cl^-$).

• Boiling Point Trends:     * Stronger IMFs result in higher boiling points because more energy (heat) is required to overcome the electrostatic forces holding liquid molecules together.     * Pentane vs. Ethane: Pentane ($CH_3CH_2CH_2CH_2CH_3$) has a higher boiling point than Ethane ($CH_3CH_3$) because it is a larger molecule with more electrons and greater surface area, leading to stronger LDF.     * Nonane vs. 2,3,4-trifluoropentane: Although they have similar molar masses, Nonane has a higher boiling point because its straight-chain structure provides a larger surface area for LDF compared to the more compact trifluoropentane.

Thermochemistry: Bond Energy and Reaction Steps

• Energy and Bonds:     * Bond Breaking: Always an endothermic process (requires energy absorption).     * Bond Formation: Always an exothermic process (results in energy release).     * Process Analysis: In a two-step reaction where Step 1 is $Cl_2 \rightarrow 2Cl$ and Step 2 is $Cl + CH_4 \rightarrow CH_3Cl + H$:         * In Step 1, bonds are broken and energy is absorbed.         * In Step 2, new bonds are formed and energy is released.

• Enthalpy (ΔH):     * Calculated as $\Delta H = (\text{Energy of Reactants}) - (\text{Energy of Products})$.     * For the reaction $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$, using bond energies ($H-H: 430\,\text{kJ/mol}$, $Cl-Cl: 240\,\text{kJ/mol}$, $H-Cl: 430\,\text{kJ/mol}$):         * $\Delta H = (430 + 240) - 2(430) = 670 - 860 = -190\,\text{kJ/mol}$.         * A negative $\Delta H$ indicates a decrease in energy (exothermic), where products are at a lower energy state than reactants.

Physics: Kinematics and Motion

• Constant Acceleration Equations:     * Velocity: $v_f = v_i + at$     * Displacement: $d = v_i t + \frac{1}{2}at^2$     * Velocity-Displacement: $v_f^2 = v_i^2 + 2ad$

• Practice Problems:     * Car Acceleration: A car at $12.0\,\text{m/s}$ accelerates at $2.5\,\text{m/s}^2$ for $6.0\,\text{s}$.         * $d = (12.0)(6.0) + 0.5(2.5)(6.0)^2 = 72 + 45 = 117\,\text{m}$.     * Jet Landing: A jet lands at $71.4\,\text{m/s}$ and stops in $949\,\text{m}$.         * $0^2 = (-71.4)^2 + 2a(949)$.         * $a = \frac{-(71.4)^2}{2(949)} = -2.69\,\text{m/s}^2$. The magnitude is $2.69\,\text{m/s}^2$ and the direction is North (opposite to South travel).

• Projectile Motion:     * Horizontal velocity ($v_x$) remains constant if friction is neglected.     * Vertical velocity ($v_y$) changes due to gravity ($g = -9.81\,\text{m/s}^2$).     * Example: Initial $v_x = 40\,\text{m/s}$, $v_y = 30\,\text{m/s}$. After $2.0\,\text{s}$:         * $v_x = 40\,\text{m/s}$.         * $v_y = 30 + (-9.81)(2.0) = 30 - 19.62 = 10.38\,\text{m/s}$.     * Launch Angles: Launching at $45^{\circ}$ provides the furthest horizontal distance. Launching at a higher angle (e.g., $60^{\circ}$) increases air time but decreases horizontal distance compared to $45^{\circ}$.

Physics: Forces, Inclined Planes, and Gravity

• Newton's Second Law: $F_{net} = ma$.

• Inclined Planes:     * Force of gravity parallel to the incline: $F_{||} = mg \sin(\theta)$.     * Force of gravity perpendicular to the incline: $F_{\perp} = mg \cos(\theta)$.     * Example on Acceleration: A box of mass $m$ slides up a ramp with velocity $+v$ and kinetic friction $f$.         * The forces acting against the motion (down-ramp) are friction ($-f$) and the parallel component of weight ($-mg \sin \theta$).         * Equation: $-f - mg \sin \theta = ma$.

• Crate on Horizontal Surface with Angled Cable:     * $m = 8\,\text{kg}$, $\theta = 40^{\circ}$, $\mu_k = 0.25$, $F_N = 50.35\,\text{N}$.     * Friction force: $f_k = \mu_k F_N = 0.25 \times 50.35 = 12.59\,\text{N}$.     * At constant speed, horizontal forces are balanced: $F_{applied, x} = f_k$.     * $F_A \cos(40^{\circ}) = 12.59\,\text{N}$.     * $F_A = \frac{12.59}{\cos(40^{\circ})} \approx 16.43\,\text{N}$.

• Universal Gravitation:     * Formula: $F = G \frac{m_1 m_2}{r^2}$.     * If a satellite's distance increases from the surface (radius $R$) to three radii above the surface ($3R$), the total distance from the center is $R + 3R = 4R$.     * New force: $F_{new} = \frac{G m_1 m_2}{(4R)^2} = \frac{1}{16} \frac{G m_1 m_2}{R^2} = \frac{1}{16}F$.

Electricity and Circuits

• Equivalent Resistance ($R_{eq}$):     * Series: $R_{eq} = R_1 + R_2 + R_3 + ...$     * Parallel: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$     * Circuit Comparisons:         * Circuit A (three $3\,\Omega$ in series): $R_{eq} = 9\,\Omega$.         * Circuit B (three $3\,\Omega$ in parallel): $R_{eq} = 1\,\Omega$.         * Circuit C (three $4\,\Omega$ in series): $R_{eq} = 12\,\Omega$.         * Circuit D (four $4\,\Omega$ in parallel): $R_{eq} = 1\,\Omega$.         * Smallest resistance is found in parallel circuits with many branches.

Experimental Data and Molarity

• Concentration Calculations:     * Molarity: $M = \frac{\text{moles of solute}}{\text{Liters of solution}}$.     * Example: $0.25\,\text{moles}$ in $250\,\text{mL}$.         * $M = \frac{0.25\,\text{mol}}{0.25\,\text{L}} = 1.0\,\text{M}$.

• Gravimetric Analysis (Precipitation):     * Scenario: $BaCl_2 + Na_2SO_4 \rightarrow BaSO_4(s) + 2NaCl$.     * Mass of empty watch glass: $24.15\,\text{g}$.     * Final dried mass of watch glass and precipitate: $25.31\,\text{g}$.     * Mass of $BaSO_4 = 25.31 - 24.15 = 1.16\,\text{g}$.     * Moles of $BaSO_4 = \frac{1.16\,\text{g}}{233.4\,\text{g/mol}} \approx 0.005\,\text{mol}$.     * Ratio of $Ba^{2+}$ to $BaSO_4$ is $1:1$, so solution contained $0.005\,\text{mol}$ in $25.0\,\text{mL}$.     * Concentration: $\frac{0.005\,\text{mol}}{0.025\,\text{L}} = 0.2\,\text{M}$.

Questions & Discussion

• Q: Why does distilled water have a higher surface tension than vegetable oil or ethanol?     * A: Distilled water forms more and stronger hydrogen bonds than the other liquids, resulting in higher surface tension that allows more drops to sit on a penny.

• Q: What happens to a paperclip floating on water when soap is added?     * A: Dish soap molecules (surfactant) decrease the surface tension of the water by disrupting the hydrogen bonds, causing the paperclip to sink.

• Q: Why do regions near large lakes have smaller temperature changes than deserts?     * A: Water has a high specific heat capacity, meaning it absorbs and releases large amounts of heat with minimal temperature change compared to sand.

• Q: How does the lipid membrane of a cell act as a barrier?     * A: Lipids are nonpolar (hydrophobic) and do not mix with water, which maintains the separation between the interior and exterior aqueous environments.