Reading...
Play button
Play button
Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
23 Cards in this Set
- Front
- Back
|
Single Covalent Bonds
|
One pair of electrons is shared
|
|
|
Double Covalent Bonds
|
Two pairs of electrons are shared (typically C, N, O, S )
|
|
|
Triple Covalent Bond
|
Three pairs of electrons are shared ( typically C or N)
|
|
|
If atoms need 1 electron
|
it will usually form 1 covalent bond - H and Halogen (group 17) will form only one bond.
|
|
|
If atoms need 2 electron
|
it will usually form 2 covalent bonds ( group 16 )
|
|
|
If atoms need 3 electron
|
it will usually form 3 covalent bonds ( group 15)
|
|
|
Bond length
|
the distance between bonded nuclei - determined by size of bonding atoms & how many electrons are shared
* as more electrons are shared, bond length increases |
|
|
Shorter bond length
|
= stronger bond
|
|
|
Which bond type is weakest?
|
Single ( longest )
|
|
|
Which bond type is stronges?
|
Triple ( shortest )
|
|
|
EN difference and bond character
|
>1.7 = mostly ionic
0.4-1.7 - polar covalent <0.4 = nonpolar covalent |
|
|
Polar Molecules always have
|
1. At least one polar covalent bond.
2. One slightly positive side and one slightly negative side |
|
|
Synthesis Reactions
|
2 or more reactants join to form a single product
|
|
|
Decomposition Reaction
|
A single reactant breaks apart
|
|
|
Combustion
|
Occurs when one of the reactants is oxygen
(always Carbon Dioxide and Water) |
|
|
Single Replacement Reactions
|
One element replaces the atoms of another element in a compound
|
|
|
Double Replacement Reactions
|
Involves an exchange of ions between two compounds
|
|
|
Calculating % Compostion
|
1. Find the total molar mass of each element in the compound
2. Find the molar mass of the entire compound 3. Divide the total molar mass of each element by the molar mass of the compound then multiply by 100 4. Check that all your %'s add up to 100 |
|
|
Calculating Empirical Formula
|
1. Change % to g ( assume 100g of compound so 30% = 30 g)
2. Convert Element from g to moles 3. Divide each mole amount by the smallest number from step 2 4. Change to a whole number = subscript in empirical formula |
|
|
Calculating Molecular Formula
|
1. Calculate the mass of the empirical formula
2.Divide molar mass of compound (will be given ) by mass of empirical formula 3. Multiply each subscript in the empirical formula by the answer from step 2 |
|
|
Limiting Reactant
|
reactant used up first ( you run out of it)
* Determines the amount of product that can be formed |
|
|
Excess Reactant
|
the other reactant ( not used up )
|
|
|
Determining Limiting Reactant
|
1. Perform conversion for each reactant= amount needed
2. Compare amount needed with amount present 3. Whichever doesnt have enough = limiting |
