You need to find which intermolecular forces are between the molecules
dipole-dipole,h bonds, etc.
I'm not very good at explaining but this is what my prof said to help us
Identify the class of the molecule or molecules you are given. Are they nonpolar species, ions or
do they have permanent dipoles? Is there only one species or are there two?
In the case of ONE species (i.e., a pure substance), the intermolecular forces will be between
molecules of the same type. So if you are dealing with ions, the intermolecular forces will be ION-
ION or IONIC. If you are dealing with dipoles, then the intermolecular forces will be DIPOLE-
DIPOLE. If you are dealing with nonpolar species, the intermolecular forces will be DISPERSION
or VAN DER WAALS or INDUCED DIPOLE-INDUCED DIPOLE (the last three are desciptions
of the same interaction; regrettably we cannot call them nonpolar-nonpolar!).
In the case of TWO species (i.e., a mixture), the intermolecular forces will be between molecules of
one type with molecules of the second type. For example, ION-DIPOLE interactions exist between
ions dissolved in a dipolar fluid such as water.
Answer:
The molarity of the formed CaBr2 solution is 0.48 M
Explanation:
Step 1: Data given
Number of moles CaBr2 = 0.72 moles
Volume of water = 1.50 L
Step 2: Calculate the molarity of the solution
Molarity of CaBr2 solution = moles CaBr2 / volume water
Molarity of CaBr2 solution = 0.72 moles / 1.50 L
Molarity of CaBr2 solution = 0.48 mol / = 0.48 M
The molarity of the formed CaBr2 solution is 0.48 M
Should be 18, well i guess I have to have 20 characters so good luck
The relationship of radiation with distance obeys the inverse square law. Therefore, doubling the distance decrease the radiation by a factor of 4. The new count is 250.
1) Applying the same principle, the count decreases by a factor of 100. The new count is 10
2) An alpha particle is 4He2 and the Hydrogen can be represented as 1H1
14N7 + 4He2 - 1H1
= 17X8
Proton number 8 belongs to Oxygen. Therefore, the resultant nucleus is:
17O8
3) 185Au79 - 4He2
= 181Ir77
4) X - 4He2 = 234Th90
X = 238U92
5) Beta emission results in the same nucleon number but an increase in the proton number; therefore, the result is:
234Pa91
Answer:
can oxygen exist as a liquid and solid
