Physical Science Reviewer - 12-Aristotle - Q2 Quiz 1 PDF

**Physical Science Reviewer** **12-Aristotle 2^nd^ Quarter -- Quiz 1** **CATALYST** - A catalyst is a substance which speeds up a reaction, but is chemically unchanged at its end. - When the reaction has finished, the mass of catalyst is the same as at the beginning. - It changes...

**Physical Science Reviewer** **12-Aristotle 2^nd^ Quarter -- Quiz 1** **CATALYST** - A catalyst is a substance which speeds up a reaction, but is chemically unchanged at its end. - When the reaction has finished, the mass of catalyst is the same as at the beginning. - It changes the equilibrium constant of a reaction. **ACTIVATORS -** *Positive Catalyst* **INHIBITORS** *- Negative Catalyst* **MnO₂** Manganese Dioxide **H₃PO₄** Phosphoric Acid **Fe** Iron **H₂SO₄** Sulfuric Acid **V₂O₅** Vanadium Pentoxide **Alcohols** Methanol, Ethanol, Etc. **Haber's Process**action between magnesium and oxygen to form magnesium oxide (MgO\\text{MgO}MgO). She starts with 15.5 g15.5 \\, \\text{g}15.5g of magnesium (Mg\\text{Mg}Mg) and 20 L20 \\, \\text{L}20L of oxygen gas (O2\\text{O}\_2O2​) at standard temperature and pressure (STP). **2Mg + O₂ → 2MgO** 2Mg = 15.5g O = 20L (STP)A scient - The **Haber process** uses an **iron catalyst** to produce ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. - **Iron Catalyst** provides a surface where nitrogen and hydrogen molecules can adsorb and react. It facilitates the breaking of the strong triple bond in nitrogen (N≡N), allowing it to combine with hydrogen more readily. - This catalyst allows the reaction to occur efficiently at high temperatures (400-500°C) and pressures (150-250 atm) **N2(g)+3H2(g) ---**Fe Catalyst**→ 2NH3(g)** **CHEMICAL REACTIONS** **Chemical Equation** - Is the shorthand way of expressing chemical reaction using numbers, symbols and formula. - It has two parts: **Reactants** and **Products** **Reactants** - These are the substances that changes when it is combined with another substance in a chemical reaction - The starting materials **Products** - The new substance - The substances that result of the chemical reaction In a chemical equation, you will see two numbers. A big one on the left and a small one on the right. These are called **Coefficient** and **Subscript**. **Coefficient** - Is a number before the symbol or formula of a substance. - **6**CO₂ + **6**H₂O → C₆H₁₂O₆ + **6**O₂ *(photosynthesis)* **Subscript** - Is the number of atoms of the element found in the substance. - 6CO**₂** + 6H**₂**O → C**₆**H**₁₂**O**₆** + 6O**₂** *(photosynthesis)* **Symbols used in Chemical Equations** +-----------------------------------+-----------------------------------+ | **SYMBOL** | **MEANING** | +===================================+===================================+ | \+ | Indicates that the substances are | | | **added** or mixed. | | | | | | Can be read as "**reacts with**" | | | when used in reactant, read as | | | "**and**" when it is used in | | | product. | +-----------------------------------+-----------------------------------+ | → | Means "**yields**" or | | | "**produces**" separates the | | | reactants from the products. | +-----------------------------------+-----------------------------------+ | ↓ | Means that a precipitate is | | | formed. | +-----------------------------------+-----------------------------------+ | ↑ | Indicates that a gas has evolved. | +-----------------------------------+-----------------------------------+ | (s) | Designates that the reactants or | | | products are in **solid** form. | +-----------------------------------+-----------------------------------+ | (l) | Designates that the reactants or | | | products are in **liquid** form. | +-----------------------------------+-----------------------------------+ | (g) | Designates that the reactants or | | | products are in **gas** form. | +-----------------------------------+-----------------------------------+ | (aq) | Designates an aqueous solution; | | | the substance is dissolved in | | | water. | +-----------------------------------+-----------------------------------+ | Δ | Indicates that heat is supplied. | +-----------------------------------+-----------------------------------+ **Example** **REACTANTS** ***yields to*** **PRODUCT** ----------------------------- --------------------------------- ------------------------- **6CO₂(g) + 6H₂O(l)** **→** **C₆H₁₂O₆(s) + 6O₂(g)** ***Breakdown of Equation*** **6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂** **6, 6, 6** Coefficient **2, 2, 6, 12, 6, 2** Subscript **(g), (l), (s), (g)** Symbol **Types of Chemical Reaction** ***Combination or Synthesis*** - A reaction when two or more elements combine to form a single product. ***Decomposition*** - A single reactant breaks down into simpler ones. It is the opposite of combination reaction. ***Single Displacement/Replacement*** - A substance; it can be an element or a compound; capable of replacing one of the atoms of a given compound. ***Double Displacement/Replacement*** - A reaction in which ions get exchanged between two reactants, resulting to the formation of a new compound. **BALANCING CHEMICAL EQUATIONS** **Law of Conservation of Mass** - The sum of the masses of the products is always the same as the sum of the masses of the reactants. - states that mass is neither created nor destroyed in any ordinary chemical reaction. **How to Balance Equations** **Step 1** Identify the elements that need to be balanced **Step 2** Identify their subscripts *Note: If there is no number, the subscript is automatically one (1).* **Step 3** Write them down. The reactants on the left, the products on the right. **Step 4** Find a coefficient wherein when it is multiplied with the subscripts, it balances. *Note: If there is **no plus** sign in the middle, you cannot add a coefficient there.* *If a coefficient is added, all subscripts are multiplied with it if there is no plus sign.* *If there is a plus sign, you can add a coefficient on the element.* ***Easy - No Double Elements*** **Na + Cl₂ → Cl + Na₂** Na = 1 x 2 **2** Na = 2 x 1 **2** Cl = 2 x 1 **2** Cl = 1 x 2 **2** **Answer: 2Na + Cl₂ → 2Cl + Na₂** ***Easy - One Double Elements*** **Na + Cl₂ → ClNa₂** Na = 2 x 2 **4** Na = 1 x 4 **4** Cl = 1 x 2 **2** Cl = 2 x 1 **2** **Answer: 2Na₂ + Cl → Cl₂Na** ***Moderate -- One Double Element*** **C₄H₁₀ + O₂ → CO₂ + H₂O** C = 4 x 2 8 C = 1 x 8 8\ H = 10 x 2 20 H = 2 x 10 20\ O = 2 x 13 26 O(1) = 2 x 8 = 16 O(2) 1 × 10 = 10 26 **Asnwer: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O** ***\ *** ***Moderate -- Multiple Double Elements*** **NaOH + H₂CO₃ → Na₂CO₃ + H₂O** Na = 1 x 2 **2** Na = 2 x 1 **2** O(1) = 1 x 2 = 2 O(2) = 3 x 1 = 3 **5** O(1) = 3 x 1 = 3 O(2) 1 x 2 = 2 **5** H(1) = 1 x 2 = 2 H(2) = 2 x 1 = 2 **4** H = 2 x 2 **4** C = 1 x 1 **1** C = 1 x 1 1 **Answer: 2NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O** When an equation has an element inside a parenthesis, it means that the element inside is multiplied to the subscript. **Step 1** Identify the subscript outside the parenthesis **Step 2** Identify the subscript of the elements inside the parenthesis **Step 3** Multiply the subscript of elements to the subscript outside the parenthesis. *Note: You must first multiply the subscripts inside to the coefficient before distributing the outside.* **Step 4** You can write (optional) another set of elements if they are doubled to minimize confusion. ***Hard -- Distribution of Elements*** **Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O** *REACTANTS* Al = 1 x 2 2 O(1) = 1 x 2 (3) = 6 O(2) = 4 x 3 = 12 18 H(1) = 1 x 2 (3) = 6 H(2) = 2 x 3 = 6 12 S = 1 x 3 3 *PRODUCTS* Al = 1 x 2 2 O(1) = 4 x 1 (3) = 12 O(2) = 1 x 6 = 6 18 H = 2 x 6 12 S = 1 x 1 (3) 3 **Answer: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O** See that with this example, there are elements inside parenthesis. The subscripts of the elements inside are first multiplied with their respective coefficient before they are multiplied (distributed) with the coefficient outside the parenthesis. **2Al(OH₂)₃** O and H is in the parenthesis; O = 1 x 2 = 2 x 3 = 6 H = 2 x 2 = 4 x 3 = 12 We first get the product of coefficient and subscript before distributing **Tip for Balancing** - You can just write the elements in order then put their subscript beside them. After that, you can find the smallest number you can multiply with them so that they can be balanced. It's a matter of trial and error. - If there are double elements, note that when you put a coefficient, it also affects other elements. (2NaO**₂ =** 2 was added, so O**₂** is affected. It will equal to Na = 2 and O = 4) - You cannot put coefficients in between elements if there is no plus sign. (2Na4O**₂** = **Bad** 2NaO**₂** = **Good** 2Na **+** 4O**₂ = Good**) **STOICHIOMETRY** - **Stoichiometry** is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. - In Greek,** *stoikhein ***means element and ***metron** *means **measure**, so stoichiometry literally translated means the **measure of elements.** **Reactant/s or Reagent/s** - **Limiting Reactant/Reagent** What is **used up** in a reaction - **Excess Reactant/Reagent** What is **left over** in a reaction **Types of Conversion in Stoichiometry** - **Mol. to Mol.** - mol. Substance A → mol. Substance B - 1 Step Process - Need to identify the molar ratio - **Mol. to Grams or Grams to Mol.** - mol. Substance A → grams Substance B or grams Substance A → mol. Substance B - 2 Step Process - **Grams to Grams** - grams Substance A → grams Substance B - 3 Step Process **Molar vs Molecular Mass** - **Molar Mass** = One element only - Molar Mass = Atomic Mass + g/mol - Mg Atomic Mass = 24 - Mg Molar Mass = 24 g/mol - **Molecular Mass** = Whole substance - Add molar mass of two elements - Mg = 24 g/mol + O = 16 g/mol. = 40 g/mol. **Example:** *Note: Balance an equation first before solving for the stochiometric problem* *Note: Ito lang 'yung mayroon ako na example sa lecture/notes at kung ano 'yung naaalala ko.* - A chemistry student is conducting an experiment where magnesium reacts with oxygen to produce magnesium oxide **(MgO)**. The student weighs out **15.5g** of magnesium **(Mg)** for the reaction. How many grams of magnesium oxide **(MgO)** will be produced if all the magnesium reacts completely with oxygen? - Mg to MgO **Equation:** 15.5 g Mg [*x*]{.math.inline} [\$\\frac{1\\ \\text{mol\\ Mg}}{24\\text{\\ g\\ Mg}}\\text{\\ x\\ }\\frac{2\\ \\text{mol\\ Mg}O}{2\\ mol\\ Mg}\\text{\\ x\\ }\\frac{40\\ g\\ MgO}{1\\ mol\\ MgO}\\text{\\ \\ }\$]{.math.inline}= 25.83 g MgO **How to Solve:** **Step 1: Analyze** **2Mg + O₂ → 2MgO** (Grams to Grams) *Note: The coefficient here is no. of moles.* 2Mg = 15.5g *Note: For every 2 mol Mg, 2 mol MgO is created* O = 20L (STP) *Note: For every 1 mol O, 2 MgO is Created* 2MgO = ? from Mg **Step 2: Convert Mg Grams to Moles** - Find the molar mass of the element (Mg) - Molar Mass = Atomic Mass + g/mol - Mg Atomic Mass = 24 - Mg Molar Mass = 24 g/mol - Divide (Round off to nearest 4) **Moles of Mg =** [\$\\frac{\\text{mol\\ Mg}}{\\text{Molar\\ Mass\\ of\\ Mg}}\$]{.math.inline} = [\$\\frac{15.5\\ g}{24\\ g/mol}\$]{.math.inline} = 0.6458 mol Mg **This Becomes:** 15.5 g Mg [*x*]{.math.inline} [\$\\frac{1\\ mol\\ Mg}{24\\ g\\ Mg}\$]{.math.inline} **Step 3: Use the Mole Ratio from the Balanced Equation (2Mg + O₂ → 2MgO)** - For every 2 mol Mg, 2 mol MgO is created - This simplifies to 1:1, meaning **2 mol Mg** produces **2 mol MgO.** **Ratio from Equation =** [\$\\frac{2\\ mol\\ MgO}{2\\ mol\\ Mg}\$]{.math.inline} **=** 1 **Step 4: Find the Molecular Mass of Substance Produced (2MgO)** - Find Molecular Mass of MgO - Add the atomic mass of Mg and atomic mass of O - Mg = 24 + O = 16 = MgO = 40 - Then add g/mol (MgO = 40 g/mol) - Write down the ratio below the Molecular Mass solved from the step above **Molecular Mass of MgO =** [\$\\frac{40\\ g\\ MgO}{1\\ mol\\ MgO}\$]{.math.inline} **=** 40 **\ ** **Step 5: Answer (25.83g MgO)** - To answer, divide all the parts that need to be divided. - 15.5 g Mg [*x*]{.math.inline} [\$\\frac{1\\ mol\\ Mg}{24\\ g\\ Mg}\$]{.math.inline} = **0.6458** - [\$\\frac{2\\ mol\\ MgO}{2\\ mol\\ Mg}\$]{.math.inline} = **1** - [\$\\frac{40\\ g\\ MgO}{1\\ mol\\ MgO}\$]{.math.inline} = **40** - Multiply all the products - **0.6458 x 1 x 40 = 25.832** - Round off to the nearest 2 decimal points - 25.832 = **25.83** - Therefore, the mass of MgO in this calculation is **25.83 MgO** **Example:** - A laboratory experiment is being conducted where magnesium reacts with oxygen to produce magnesium oxide **(MgO).** The reaction begins with **20L** of oxygen gas (**O₂**​) at standard temperature and pressure **(STP - 22.4 L)**. How many grams of magnesium oxide **(MgO)** will be produced if 20L of oxygen reacts completely with magnesium? - O to MgO **Equation:** 20 L O₂ [*x*]{.math.inline} [\$\\frac{1\\ mol\\ O₂}{22.4\\ L\\ O₂}\\text{\\ x\\ }\\frac{2\\ mol\\ MgO}{1\\ mol\\ O₂}\\text{\\ x\\ }\\frac{40\\ g\\ MgO}{1\\ mol\\ MgO}\\text{\\ \\ }\$]{.math.inline}= 71.43 g MgO Step 1: **2Mg + O₂ → 2MgO** (Grams to Grams) *Note: The coefficient here is no. of moles.* 2Mg = 15.5g *Note: For every 2 mol Mg, 2 mol MgO is created* O = 20L (STP) *Note: For every 1 mol O, 2 MgO is Created* 2MgO = ? from O **Step 2: Convert Mg Grams to Moles** - Find the molar mass of the element (O) - Molar Mass = Atomic Mass + g/mol - O Atomic Mass = 16 - O Molar Mass = 16 g/mol - Divide (Round off to nearest 4) **Moles of O₂ =** [\$\\frac{\\text{Mass}\\ O₂}{\\text{Molar\\ Mass\\ of\\ }O₂}\$]{.math.inline} = [\$\\frac{20\\ L}{22.4\\ L\\ O₂}\$]{.math.inline} = 0.8929 mol O₂ **This Becomes:** 20 L O₂ [*x*]{.math.inline} [\$\\frac{1\\ mol\\ O₂}{22.4\\ L\\ O₂}\$]{.math.inline} **Step 3: Use the Mole Ratio from the Balanced Equation (2Mg + O₂ → 2MgO)** - For every 2 mol Mg, 2 mol MgO is created - This simplifies to 1:1, meaning **2 mol Mg** produces **2 mol MgO.** **Ratio from Equation =** [\$\\frac{2\\ mol\\ MgO}{1\\ mol\\ O₂}\$]{.math.inline} **=** 2 **Step 4: Find the Molecular Mass of Substance Produced (2MgO)** - Find Molecular Mass of MgO - Add the atomic mass of Mg and atomic mass of O - Mg = 24 + O = 16 = MgO = 40 - Then add g/mol (MgO = 40 g/mol) - Write down the ratio below the Molecular Mass solved from the step above **Molecular Mass of MgO =** [\$\\frac{40\\ g\\ MgO}{1\\ mol\\ MgO}\$]{.math.inline} = 40 **Step 5: Answer (71.43g MgO)** - To answer, divide all the parts that need to be divided. - 20 L O₂ [*x*]{.math.inline} [\$\\frac{1\\ mol\\ O₂}{22.4\\ L\\ O₂}\$]{.math.inline} = **0.8929** - [\$\\frac{2\\ mol\\ MgO}{1\\ mol\\ O₂}\$]{.math.inline} **= 2** - [\$\\frac{40\\ g\\ MgO}{1\\ mol\\ MgO}\$]{.math.inline} = **40** - Multiply all the products - **0.8929 x 2 x 40 = 71.432** - Round off to the nearest 2 decimal points - 71.432 = **71.43** - Therefore, the mass of MgO in this calculation is **71.43 MgO** **Final Answers:** 2MgO from 2Mg = **25.83 MgO** Moles of 2Mg = **0.6458mol Mg** 2MgO from O₂ = **71.43 MgO** Moles of O₂ = **0.8929mol O₂** **Limiting and Excess Reactants** - You have **0.6458 mol Mg** and **0.8929 mol O2​.** - The **Lesser number** is the **Limiting Reactant** (**0.6458)** - The **Greater number** is the **Excess Reactant** (**0.8929)** *Tip: greater number of moles means it has more possibility to not use up all of the moles, meaning that the greater number of moles usually means that element will have excess, hence excess reactants.* *Goodluck sa quiz! Papasa rin tayo! Sana naman i-crushback na ako nung isa riyan...* *Padayon, guys! Kaya natin 'to! Bibili na niyan tayo ng bibingka* *J.E.P*