{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chemistry Mid-Term 2 Notes

# Chemistry Mid-Term 2 Notes - ViCi = VfCf Strong...

This preview shows pages 1–2. Sign up to view the full content.

ViCi = VfCf Strong electrolytes are good conductors of electicity To find the molecular formula of 2 given masses: o Convert grams into mols of given compounds o Divide the one with more mols, by the one with less mols This tells you how many of each compound are in the emp. Form. o Find the Empirical Formula [(grams/mol)+(grams/mol) = Empirical Formula Mass o To find molecular formula: [(Molar mass of compound)/(emp. Formula mass)] = # Formulas per mol o Now multiply emp. Formula by # Formulas per mol (Avg. Amu)=(%ele.2)(Amu ele2) + (100-%ele2)(A) Solve for A To find # of atoms o Find total molar mass o Convert grams to mols o Multiply into avogadro’s mol o Turn into atoms Calculating Mass % o Divide the mass of the element you’re trying to find % of By the molar mass of the compound it’s in Finding empirical formulas o Convert each element to mols o Divide each mol by the smallest # Finding pressure o [Mols(nR)(T)]/V o Avg. KE = (3/2)(RT) o Root Mean Square Velocities (avg. velocity of gas particle) = (3RT/M)^(1/2) Finding concentrations o Concentration = amount of X / Vsoln o Convert compound from g mols o Divide my volume of solution in Liters Force = Mass x Acceleration (F=MA) o Acceleration is due to gravity SI units of force is Newton o 1 newton = 1kgm/s^2 also the force unit Pressure = force/area SI unit of pressure is Pascal (PA), 1 PA = 1kg/ms^s = 1n/m^2 o Standard Atmosphere = 101,325 PA= 1 atm = 760 torr = 760 mmhg Boyle’s Law o

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Chemistry Mid-Term 2 Notes - ViCi = VfCf Strong...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online