Chemistry Exam Notes

Types of Reactions

• The student should study the types of reactions to understand each one.
• Single Replacement:
  • A single element and a compound react.
  • Example: Fe + NaBr3 \rightarrow FeBr3 + Na
  • After the reaction, there will be a different single element.
  • In the example, iron (Fe) forms a positive ion, so it will be the single element on the product side.
  • Sodium (Na) and Bromide (Br) form the new compound, Sodium Bromide (NaBr).
  • Sodium has a +1 charge, and Bromide has a -1 charge, so they combine directly.
• Balancing Equations:
  • Balance chemical equations using coefficients.
  • Balance with coefficients: Adjust the numbers in front of the compounds to ensure the number of atoms of each element is equal on both sides of the equation.
  • The example equation is unbalanced: Fe + NaBr3 \rightarrow FeBr3 + Na
  • Original count (reactants): 1 Fe, 3 Na, 3 Br
  • To balance, start with the element that is not balanced (e.g., Na).
  • Place a coefficient of 3 in front of NaBr: Fe + 3NaBr3 \rightarrow FeBr3 + 3Na
  • Now, balance the remaining elements by inspection.

Double Replacement Reactions

• Double Replacement:
  • Two compounds react, and the positive ions switch places.
  • Example: NaOH + H2SO4 \rightarrow Na2SO4 + H_2O
  • Sodium (Na) combines with Sulfate (SO4), and Hydrogen (H) combines with Hydroxide (OH).
  • Sodium has a +1 charge, and Sulfate has a -2 charge, so they combine as Na2SO4.
  • Hydrogen and Hydroxide combine to form water (H_2O or HOH).
• The general process is:
  • Identify the positive and negative ions in the reactants.
  • Swap the positive ions and combine them with the negative ions from the other reactant.
  • Make sure the resulting compounds are neutral by balancing the charges.
  • The number of atoms in the products is calculated based on the charges (oxidation numbers) of the new molecule.

Balancing Double Replacement Reactions

• Example: NaOH + H2SO4 \rightarrow Na2SO4 + H_2O
  • Reactant count: 1 Na, 1 OH, 2 H, 1 SO_4
  • Product count: 2 Na, 1 SO_4, multiple H and O in water.
  • Place a 2 in front of NaOH to balance Sodium: 2NaOH + H2SO4 \rightarrow Na2SO4 + H_2O
  • Place a 2 in front of H2O to balance Hydrogen and Hydroxide: 2NaOH + H2SO4 \rightarrow Na2SO4 + 2H2O

Key Principles for Predicting Products

• Positive Ion:
  • The positive ion from one reactant combines with the negative ion from the other reactant.
• Charge Balance:
  • Ensure the compounds formed are neutral by properly balancing the charges.
• Subscripts:
  • Subscripts on the reactant side do NOT determine the subscripts on the product side.
• Balancing:
  • Balancing is based on making sure the compounds are neutral.
  • Balancing ensures there's no net charge in the compound.
• Underlying Knowledge:
  • This unit builds upon the previous unit.
  • A solid understanding of the previous unit is essential for predicting products.
  • Example: PbSO4 + AgNO3 \rightarrow Pb(NO3)2 + Ag2SO4
    • Lead (Pb) has a +2 charge because Sulfate (SO_4) has a -2 charge.
    • Silver (Ag) has a +1 charge, and Nitrate (NO_3) has a -1 charge.
    • Ensure charges remain consistent on both sides of the reaction.

Oxidation Numbers

• Example: MnO2 + Sn(OH)4 \rightarrow Mn(OH)4 + SnO2
  • Manganese (Mn) has a varying oxidation number, so it must be figured out.
  • Oxygen has a -2 charge, thus 2 oxygen atoms make it 4 minuses, which means Mn is +4 to balance it.
  • Sn(OH)₄: Hydroxide (OH) has a -1 charge, and there are four of them, so 10 is +4 to balance it.
  • Mn(OH)₄: The 4 hydroxides give it 4 minuses so Mn is +4 to balance it.
  • The Oxidation Is Potential:
    • +1,+2, skip the DIP, +3, +/4 (plus or minus), -3, -2, Oxygen, -1, -7
    • The oxidation number does not change when it goes to the product side.

Balancing Combustion Reactions

• Combustion Reactions:
  • Involve a hydrocarbon (a compound containing carbon and hydrogen) or a compound containing carbon, hydrogen, and oxygen reacting with oxygen.
  • Products are always carbon dioxide and water.
  • Example: C2H4O2 + O2 \rightarrow CO2 + H2O
• Balancing Steps:
  • Write CHO on both sides to keep track of atoms.
  • Balance carbon first, then hydrogen, then oxygen.
  • Adjust coefficients to ensure the number of atoms of each element is equal on both sides.
  • Ensure total number of oxygen atoms matches on both sides.
    • If you have C5H{12}O2 + O2
      • You will get CO2 and H2O EVERY SINGLE TIME.
      • Match your beginning accounts to look like the step that you want to take and copy the carbons you have from the reaction side.
      • Then to balance your equation look at your changing hydrogen and your oxygens. You do not look there until you change your hydrogen.

Double Replacement Examples

• Identify positive ions: In a compound, the positive ion includes the hydrogen, potassium, and/or silver. The negative ions include the rest i.e. chlorine.
  • Mn(OH)₄ + SnO₂ is balanced with the same +4 charge because they balance already.
  • For compound plus compounds, you must have double replacement.
    • You would swap the partners from there.
  • With MnO₄ + ZnCl₂ you will have MnCl in order to correctly balance it.
    • The one balanced is: Zn + MnO₄
      • To do this you must know what the charges on MnO₄ are.
        • The easy way to remember that charge is that with one potassium, potassium has one positive charge. Which means that Mn is actually one minus.

Nuclear Chemistry

• Definition:
  • The study of changes to the nucleus of an atom.
• Radioactivity:
  • Emission of radiation.
• Radioactive or Nuclear Decay:
  • The process by which an unstable nucleus loses energy and matter by emitting radiation.
  • An unstable nucleus occurs when there are different numbers of neutrons and protons.
• Alpha Particle:
  • Symbol: H e
  • Mass number: 4
  • Atomic number: 2
• Beta Particle:
  • It has a zero and a negative one because all of that is an electron. It is the loss of an electron by an unstable nucleus.
  • Notation: e
  • Mass number: 0
  • Atomic number: -1

Nuclear Equations

• Balancing:
  • Ensure the mass numbers and atomic numbers are equal on both sides of the equation.
• Example (Beta Emission):
  • Protactinium-234 (Pa-234) undergoes beta decay.
  • Equation: Be (e) + 234 / 92 x
  • The symbol for beta is e with a zero and a negative one from the equation
    • Zero plus what equals two thirty four equals zero, giving it the 234 / -1 plus what give me 91
• Example (Alpha Emission):
  • Antimony-116 (Sb-116) emits an alpha particle.
  • Equation: Sb-116 → He + In
  • Find Antimoy on the periodic table, that gives you 116 / 52 Sb / He + In
    • Four plus what gives me 116? That is 112. From that you have two plus what gave me 51? 49.
  • From that you would look for number 49 in a new element.
• Key Points:
  • All reactions will involve emission, not absorption.
  • Focus on Alpha and Beta particles only.
  • Match and balance the sides using mass and atomic numbers.