Arrange the following compounds in order of increasing solubility in water:

* O2
* LiCl
* Br2
* CH3OH

Like dissolves like; that is, polar compounds usually are soluble in water and non-polar compounds are not soluble in water. From that description, make your choices and post what you think. We can help you from there. OR, tell us what you don't understand about the question.

So, O2 is nonpolar, and is therefore not soluble. Br2 is also nonpolar, so how do I know which one is more soluble? LiCl is ionic - does that mean it is soluble? I know NaCl is soluble and that's an ionic compound. CH3OH has some polar and nonpolar bonds. Where do I go from there?

You are right that oxygen is only slightly soluble in water. Bromine reacts with water to form HOBr and HBr so there is more solublity there. LiCl, if it is ionic, must be polar and (you did well to compare with NaCl because both Li and Na are in group IA of the periodic table) so you would expect LiCl to be quite soluble. Methyl alcohol is essentially miscible with water in all proportions (see the OH group on CH3OH and note HOH has the same OH from water). So it is an educated guess as to which, LiCl or CH3OH, has the greater solubility. I looked up LiCl and 1 gram will dissolve in 1.3 mL cold water or 0.8 mL hot water (VERY soluble). I believe CH3OH is miscible ihn all proportions so I would rank it higher. However, you form your own conclusion.

so it would be O2, Br2, LiCl, CH3OH. Very confusing.

Br2
CH3OH
LiCl
O2

What is the I- concentration just as AgCl begins to precipitate when 1.0 M AgNO3 is slowly added to a colution containing 0.020 M Cl- and 0.020 M I-?

it's based on intermolecular forces too. br and o are both nonpolar so it's all like "why would one be more soluable than the other?" but it's because br has more dispersion forces since it has more electrons. so br is a little bit more polarish (although not technically polar), and since water is very polar, that means it wants to bond with the most polarish atom. so o2 is definently first. then br. what about licl and ch3oh? well, both are polar, but ch3oh has hydrogen bonds. hydrogen bonds are stronger. water, too, has hydrogen bonds. so it wants to bond with another hydrogen bond more than it wants to bond with just a dipole. therefore ch3oh is the most. why is licl before the diatomics, though? because licl still behaves more polarish than the diatomics. just think about intermolecular forces and it all makes sense. (sorry if this is really confusing... it makes no sense without a firm grasp of intermolecular forces plus i've been studying this stuff for the past six hours -_-)

Arrange compounds in order of increase lattric energy MgCl2.SrCl2. BaCl2.MgF2. Bal2

Inceasing order of acidic strength HCl, HF HI, HBr

To determine the I- concentration when AgCl begins to precipitate, you need to compare the solubility product constant (Ksp) of AgCl and AgI. AgCl and AgI are both ionic compounds, and their solubilities can be determined using their respective solubility product constants.

The solubility product constant (Ksp) for AgCl is given by:

Ksp = [Ag+][Cl-]

Similarly, the solubility product constant (Ksp) for AgI is given by:

Ksp = [Ag+][I-]

Since you want to find the I- concentration when AgCl begins to precipitate, you need to compare the Ksp values for AgCl and AgI.

The Ksp value for AgCl is 1.77 x 10^-10, while the Ksp value for AgI is 8.52 x 10^-17.

Since AgI has a smaller Ksp value, it is less soluble than AgCl. This means that AgI will precipitate first when the ions exceed their solubility product constant.

To find the I- concentration just as AgCl begins to precipitate, you need to calculate the concentration of Ag+ when AgCl begins to precipitate, using the solubility product constant of AgCl.

Given that the concentration of Cl- is 0.020 M, the concentration of Ag+ at this point will also be 0.020 M (to satisfy the equation for the Ksp).

As a result, the concentration of I- can be calculated using the solubility product constant of AgI. Rearranging the equation for Ksp:

[Ag+][I-] = Ksp

0.020 M × [I-] = 8.52 x 10^-17

Therefore, [I-] = (8.52 x 10^-17) / 0.020 M

Calculating this value gives the concentration of I- at the point where AgCl just begins to precipitate.