Chem 105n 1st EditionExam # 2 Study GuidePRACTICE PROBLEMS AND TEST REMINDERSTypes of Radiation:- Alpha  α  ⁴₂He- Beta  ß  ⁰-₁ē- Postitron  ß-  ⁰+₁ē- Gamma Ray  ƴ  ⁰₀ƴ- Proton  ƿ  - Neutron  nAlpha Decay:- RnMass: 222Atomic #: 86- Rn  ? + ⁴₂HeMass: 222  218 + 4Atomic #: 86  84 + 2 - Resulting Element:PoMass: 218Atomic #: 84Beta Decay:- KMass: 42Atomic #: 19- K  ? + ⁰-₁ēMass: 42 42 + 0Atomic #: 19  20 + (-1)- Resulting Element:CaMass: 42Atomic #: 20- Units of radiation:- Curie (Ci) measures activity as the number of atoms that decay in one second for one gram of radium- Rad (radiation absorbed dose) measures the radiation absorbed by the tissues of the body- Rem (radiation equivalent) measures the biological damage caused by different types of radiation- Becquerel (Bq), which is 1 disintegration per second and the SI unit for radioactivity- Gray (Gy), which is defined as the joules of energy absorbed by 1 kilogram of body tissue and is equal to 100 rad- Sievert (Sv), which is the SI unit for rem, and is equal to 100 remEx. The half life of I-123 is 13 hours. How much of a 64-mg sample of I-123 remains active after 26 hours?1. Calculate the number of half lives by dividing the time (26 hours) by the half life time of I-123=22. Divide 64mg in half twice64/2 =3232/2 =16FINAL ANSWER: 16g remainingNuclear Fission:- Nucleus splits into smaller nuclei and several neutrons- Releases energyNuclear Fusion:- Combines smaller nuclei into larger nuclei- Releases energyThe Octet Rule:- When atoms form a chemical bond, it is common for them to end up with EIGHT electrons in their valence shell- Atoms form octets to by losing/gaining/sharing electrons- They are formed so that the atoms become more stable- Octets are formed when ionic/covalent bonds are formed- Ionic Bond Metal + Nonmetal - (opposites attract)- Metal atoms transfer electrons to nonmetal- Ionic Compounds are the product of ionic bonding- Covalent Bond Nonmetal + Nonmetal - (like atoms)- Nonmetals share valence electrons- Covalent Compounds are the product of covalent bondingI. Ionic Charge by Group Number:- Work to achieve electron configuration of nearest noble gas by forming positive and negative ions- Positive Ions: Have a charge equal to its group numberEx. Group 1A (1) 1+Group 1A (2) 2+Group 3A (3) 3+- Negative Ions: Obtain their charge by subtracting 8 or 18 from its group numberEx.Group 6A (16) 6-8=2-OR 16-18=2-Chemical Formula for Ionic Compounds:- Lowest whole-number ratio of ions- Sum of positively and negatively charged ions is always ZERO- Total charges are balancedTotal positive charge = Total negative chargeEx.Compute: (Na^+) + (N^3-)1) Each take each other’s charge number (Balance the charges)2) 3(+1) + 1(3-) = 0Aka(Na+) (Na+) (Na+) + (N3-) = 03) Na3NEx.Compute: (Ba2+) + (Cl-)1) Balance the charges2) (Ba2+) + (Cl-) (Cl-)= 03) BaCl2- To name a cation with TWO elements:- Identify the cation and anion- Name the cation first followed by the name of the anion- The names of transition metals with two or more positive ions (cations) use a Roman numeral after the name of the metal to identify the ion charge.- Name the anion by using the first syllable of the element name followed by “ide”Ex.- FeCl2 Iron(II)chloride- Fe2O3Iron (III) Oxide- Atoms with Greater Oxygen: end in ate (phosphate)- Atoms with Less Oxygen: end in ite (phosphite)- Atoms with a Hydrogen attached: begin with bi (bicarbonate)- Mg(NO3)2  magnesium nitrate- Cu(ClO3)2 copper(II) chlorate- Fe2(SO4)3 iron(III)sulfate- Ba3(PO3)2 barium phosphate- Aluminum nitrateAl(NO3)3- Copper(II) nitrateCu(NO3)2- Iron(III) HydroxideFe(OH)3- Tin(IV) HydroxideSn(OH)4- Potassium bromateKBrO3- Calcium carbonate CaCO3- Sodium phosphateNa3PO4- Iron(III)oxideFe2O3- Iron(II)NitriteFe(NO2)2- Ca3(PO4)2Calcium Phosphate- FeBR3Iron(III) bromide- Al2s3 aluminum sulfide- Zn(NO2)2 zinc nitrite- NaHCO3 sodium bicarbonate OR hydrogen