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when using Scientific Notation if you go <- that way what would be the sign for the answer?
It would be positive
ex of Scientific Notation: 60200000
= 6.02×10⁷
when using Scientific Notation if you go -> that way what would be the sign for the answer?
It would be negative.
ex of Scientific Notation: 0.0002077
= 2.077×10⁻⁴
Mole
6.022×10²³ (Avogadro's Number; one mole of Blah contains that many atoms of Blah)
Molar Mass
Found on The Periodic Table, ex 106.42 g/mol
Atomic Number
Found on Periodic Table; ex 1=H, 2=He
If we had 0.576 what number would we round to for our answer?
the 3rd number because there is only 3 numbers after the . and 0 never counts
Mole Equations:
1) Mass in Grams/molar mass (From periodic table)
2) # of Atoms/(6.022×10²³ Avogadro's #)
Atom Equation:
Moles × (6.022×10²³ Avogadro's #)
If you need to find a certain amount of atoms in that equation what should you do (ex. How many iron atoms does 4.44 moles of Fe₂O₃ contain?)
After you complete the equation for atoms you should then multiply the answer by the # of atoms next to the letter (ex. Fe₂ we would multiply the answer by 2)
Gram Equation:
Moles × Molar Mass (g/mol)
Formula Units Equation:
(Grams/Molar Mass) × (6.022×10²³ Avogadro's #)
Molecule Equation:
(Grams/Molar mass) × (6.022×10²³ Avogadro's number)
Molecule to Gram Equations:
1) Molecule × (Molar Mass / (6.022×10²³ Avogadro's #))
2) Molecules / (6.022×10²³ Avogadro's #) = Moles then Multiply that by molar mass = Grams
Stoichiometry Equation:
Given moles × (want # / Have #)
Note:(want = the # that we want the answer about and have = the number from which we took the moles from)
Stoichiometry Equation for Grams:
(Grams have/Molar mass) × (Want #/Have #) = grams want
(To find) Limiting Factor Equation:
Q. Once you have the limiting factor how do you find the answer of the number needed?
A (given) mol amount × (B coefficient/A coefficient) = B needed amount
Answer: once you get the limiting factor use the same equation to find the answer (use the og limiting factors mole # w/it as a)