<p>No, we do not have a choice.</p>
<p>Fe(Cl)2 is yellow and Fe(Cl)3 is orange. Right?</p>
<p>um, both are soluble, meaning you get Fe2+ which is yellow/green and Fe3+ which is red (think of rust).</p>
<p>You might be confusing those with chromate (yellow) and dichromate (orange)...</p>
<p>
[quote]
Hey, specify, did you get that Br- is first order, Bro3- is first order, but i can't figure out what H+ is...
[/quote]
</p>
<p>yes both first order. Use exp 3 and 4 for H+?</p>
<p>3.00 x 10^-3 / 3.75 x 10-4 = (.200)^y (.015) / (.100)^y (.0075)</p>
<p>I meant aqueous.</p>
<p>oh ok, thanks gfaith.</p>
<p>It totally amde this up, id on't even think it happens, but you should still eb able to balance it.
Balance Na + MnO7- ->Na+ + MnO2-</p>
<p>Make 2 half reactions (one oxidation and one reduction):</p>
<p>Reduction Half reaction:</p>
<p>MnO7- -> MnO2-</p>
<p>Balance everything that is not H or O:</p>
<p>MnO7- -> MnO2-</p>
<p>Now balance all oxygens with H20</p>
<p>MnO7- -> MnO2- + 5 H20</p>
<p>Now balance ALL hydrogens with H+. Don't forget to account for H's already in the compound (i.e. something like NaOH in redox reactions). MnO7- and MnO2- have no hydrogens so we're good here but still ahve to balance the hydrogens form water.</p>
<p>MnO7- + 10H+ -> MnO2- + 5 H20</p>
<p>Now balance out charges with electrons. Don't forget the coefficients when balancing charges (i.e. 10H+).</p>
<p>10e- + MnO7- + 10H+ -> MnO2- + 5 H20</p>
<p>You're done with this half reaction. Do the same thing for the other half reaction (Na -> Na+) to end up with:</p>
<p>Na -> Na+ + e-</p>
<p>Now you have to multiple the Na half reaction by 10 so the number of electrons doanted is the same s the amount received.</p>
<p>10 Na -> 10Na+ +10e-</p>
<p>You transferred 10 electrons.</p>
<p>Then add up and cancel reactants and products of both half reactions to come out with the entire balanced equation:</p>
<p>10Na + MnO7- + 10H+ -> MnO2- + 5H20 + 10Na+</p>
<p>If they don't tell you, assume that you are in an acidic solution and you are done. </p>
<p>10Na + MnO7- + 10H+ -> MnO2- + 5H20 + 10Na+</p>
<p>BUT, if they say basic solution, then you must add as many OH- anions to each side as there are H+ cations.</p>
<p>10Na + MnO7- + 10H+ + 10OH- -> MnO2- + 5H20 + 10Na+ + 10OH-</p>
<p>Now you must react the H+ with the OH- to form H20:</p>
<p>10Na + MnO7- + 10H20 -> MnO2- + 5H20 + 10Na+ + 10OH-</p>
<p>Now balance out the H20 to get the final answer in basic solution:</p>
<p>10Na + MnO7- + 5H20 -> MnO2- + 10Na+ + 10OH-</p>
<p>wow that is long</p>
<p>Ok, so the order of H+ is 2?</p>
<p>
[quote]
Ok, so the order of H+ is 2?
[/quote]
</p>
<p>Yes (10 characters)</p>
<p>Do you use these equations in the exam?
delta T<em>f=iK</em>f<em>molality
delta T<em>b=iK</em>b</em>molality</p>
<p>i=?</p>
<p>FeCl2 anhydrous solid is buff->no color. FeCl2 hydrate is green-blue.
FeCl3 solid is yellow.
FeCl2 (aq) is light green
FeCl3 (aq) is brown/yellow</p>
<p>i = van't hoff factor. Pretty much how many particles of solute there are when the solute is added to the solution. So for example, adding sugar to water, i = 1 because there's one molecule of sugar still. But, when adding Cu(NO3)2, i = 3 because it breaks up into one copper (II) ion and two nitrates.</p>
<p>Wow, I think you guys are overstudying descriptive chem.</p>
<p>guys, I am really worried about Beer's law.
I have never seen it before, let alone doing it in the lab.
What should I do?
Is it going to be on the FR??</p>
<p>@afruff23 you mean MnO4- right?
MnO7-?</p>
<p>Mn07- does exist</p>
<p>
[quote]
guys, I am really worried about Beer's law.
I have never seen it before, let alone doing it in the lab.
What should I do?
Is it going to be on the FR??
[/quote]
</p>
<p>Chances are, if it's going to be on the AP exam, it's going to be on the lab section of the FR. Go over spectrophotometry. It should be very easy. A=abc, and the variables are defined on the equation sheet, all you need to do (from the examples I've seen) is just plug in numbers they give you, and be able to read a graph with Absorbance vs. Concentration.</p>
<p>oh ok, thanks wxmann, I am looking at 5 on FR 2006</p>
<p>austin,
They give you a graph of wavelength vs. absorbance. Where absorbance is the highest, is where the optimal wavelength is. (see 2003 AP lab question)</p>
<p>lets just hope there's no complex ions for the number 4...if so just remember that Cu has 4 ligands (ammonia), Fe has 6 with cyonate but only one with thiocyanate...Silver has 2 ligands (for ammonia), and Zinc has four (ammonia again)</p>