7259085
Description
Flashcards by Benjamin Rogers-Newsome, updated more than 1 year ago
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Created by Benjamin Rogers-Newsome
over 7 years ago
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| Question | Answer |
| Relative Atomic Mass (Ar) | The average mass of an atom of an element, taking into account it's naturally occurring isotopes, relative to 1/12th the mass of carbon-12 |
| Relative Molecular Mass (Mr) | The average mass of a molecule compared to 1/12th the mass of carbon-12 |
| Relative Formula Mass (Mr) | The same as relative formula mass for ionic compounds as they do not exist as molecules |
| Avogadro's Constant Na | 6.022 x 10^23 The number of atoms in 12g of carbon-12 |
| A Mole | 6.022 x 10^23 particles. A substances formula mass weighed out in grams will contain one mole of that substance |
| Equation linking Moles, Mass and Mr | No. Moles n = Mass m (g) / Mr of substance |
| Equation linking Particles and Moles | No. Particles = Na x No. Moles n |
| A Solution | A solvent with solute dissolved in it |
| Equation linking Concentration, Moles, and Volume | Concentration C (moldm^-3) = No. moles n (mol) / Volume V (dm^3) |
| The ideal gas equation | Pressure P (Pa) x Volume V (m^3) = no. moles n (mol) x gas const R (JK^-1mol^-1) x temperature T (K) |
| Degrees Celcius to Kelvin | + 273 |
| KPa to pa | x 10^3 |
| cm^3 to dm^3 | x 10^-3 |
| dm^3 to m^3 | x 10^-3 |
| cm^3 to m^3 | x 10^-6 |
| m^3 to cm^3 | x 10^6 |
| m^3 to dm^3 | x 10^3 |
| dm^3 to cm^3 | x 10^3 |
| Empirical Formula | The simplest ratio of atoms of each element in a compound |
| Molecular Formula | The actual ratio of atoms of each element in one molecule of the compound. For ionic substances, actual ratio is the empirical formula. |
| State Symbols | (s) - solid (l) - liquid (g) - gas (aq) - aqueous |
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