______ molecules have a net dipole defined by the sum of individual dipoles
polar
132 dozens * (12 eggs/ 1 dozen) * ( 1 mole/ 6.02*1023 eggs)
=2.63E-21
Determine the mass of 1.27 moles of Mg(OH)2
M.W Mg(OH)2: 58.32 g/mol
(1.27 mole)*(58.32 gram/mole)74.1 g
How many electron domains are in OF2 ? Bonds? Molecular Shape?
1.000 mg * (1 g/ 1000 mg) * (1 mole/ 12.0107 g) * (6.022 x 1023 atoms / 1 mole)
=5.014 x 1022
7.46 g
The ion NO- has _____ valance electrons
Which molecule would you predict to have the greatest dipole moment?
HCl
CH4
H2
CH3OH
HF
HF
Which of the following has the greatest mass:
1 mole of Sc
1 mole of Cl2
1 mole of H2
1 mole of Ga
all have the same mass
1 mole of Cl2
If I leave 750 mL of 0.50 M sodium chloride solution uncovered on a windowsill and 150 mL of the solvent evaporates, what will the new concentration of the sodium chloride solution be?
(750 mL)*(0.50 M)=(150 mL)*(x)
0.63 M
(this is the opposite of a dilution problem but you approach the same way)
Give the name/or formula of the compound, and charge and number of each ion that makes it up
BaCO3
Iron (III) nitrate
Barium carbonate, 1 Ba2+,, 1 CO3 2-
Fe(NO3)3, 1 Fe1+ ,, 3 NO3 1-
What are the expected bond angles in a molecule of CH4
Calculate the molar mass of a compound if 0.331 mole of it has a mass of 226 g.
226 g /0.331 mol
= 683 g/mol
2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 1.50 M in concentration. How many mL are required?
(1.50 mol/L) (x) = (0.800 mol/L) (2.00 L)
x = 1.067 L
Divide the liters by 1000 to get mL, the answer is 1070 mL (notice that it is rounded off to three sig figs)
.. ..
:Cl - Cl:
.. ..
For each molecule or ion, determine the shape and indicate whether each molecule will be polar or non-polar. You must get them all right for full points:
1. CCl4
2.NH3
3. CHCl3
4. XeF2
5. BrF5
1. tetrahedral, non-polar
2.trigonal pyramidal, polar
3. tetrahedral, polar
4. linear, non-polar
5. square pyramid, polar
In the compound Cu2(OH)2CO3, which element is present in the largest percent mass?
M.W: 221.1156 g/mol
Cu: (2*63.546 g/mol) / (221.1156 g/mol) = 57.4776%
C: (1*12.0107 g/mol)/ (221.1156 g/mol) = 5.4319%
O:(5*15.9994 g/mol) / (221.1156 g/mol) = 36.1788%
H: (2*1.000794 g/mol)/(221.1156 g/mol)= 0.9117%
A 325 mL sample of solution contains 23.3 g of CaCl2
Calculate the molar concentration of Cl− in this solution
M.W CaCl2: 110.98 g/mole
23.3 g * (mole CaCl2/110.98 g) * (2 Cl/1 mole CaCl2)= 0.4199
Molarity= moles/Liters = 0.4199 moles/ 0.325 L = 1.29 M Cl-
Which compound do you expect will have a higher conductivity: NaCl or C6H12O6