<p>N2O5 + H2O -> HNO3? isnt HNO3 a strong acid? how can it form HNO3? o_o"</p>
<p>@apscared: when acid is added to the solution, it is a carbonic gas forming reaction which results in the release of CO2(g)</p>
<p>jam, a nonmetal oxide (nitrogen is non-metal) + water => acid, 2H+ 2NO3</p>
<p>I pulled it out from PR</p>
<p>@jacmoo + IAmCool: The reason why N2O5 + H2O –> HNO3 and HNO2 is because the oxidation state of nitrogen must remain the same, in N2O5 the oxidation state of nitrogen is +5, in HNO3 the oxidation state of nitrogen is also +5. However, in HNO2, the oxidation state of nitrogen is +3. So it must yield HNO3. </p>
<p>@apscared: First you want to see what all these solids will do in solution. Upon observation you can tell that all of them will disassociate in solution. When acid is added, we need to have a reaction occur in order for gas to be liberated. Lets go through all of the choices, a acid will break up into a conjugate base and a hydrogen ion in solution. So for the first choice, H+ + NO3- does not do anything. H + OH- – > H20, which is not a gas last time I checked. H+ HSO4- does not create a gas. H+ + Cl- is just H+ + Cl-. However when look at choice B, H+ + CO32- –> H2CO3. At first this may seem like nothing happens, but if you look closely we find this acid to be known as carbonic acid. Carbonic acid does not exist as H2CO3 in solution, it breaks up into H20 and CO2. Are any of these products a gas? Yes. CO2 is the gas that is liberated and thus the answer must be B. Hope this helped.</p>
<p>@lemone: make sure you know why its not HNO2 and why it is HNO3. Not sure if PR covers that. Because sometimes it is kind of tricky to find what acid is created from the non metal oxide in water.</p>
<p>apscared:</p>
<p>K2CO3 is a salt of cation of a a strong base (KBr) and an anion of a weak acid (H2CO3). This yields a basic salt.</p>
<p>^
@ Wongton, minh, and NewAccount: Thanks for the explanations!</p>
<p>@apscared: sorry forgot to mention why its basic because K2CO3 dissolved in water, two reactions</p>
<p>Just a rough sketch not balanced </p>
<p>K2 + H2O -> KOH + H+ <- strong base, no reaction occurs
CO3 + H2O -> HCO3 + OH- <-weak acid, and OH- is released as a byproduct of the reaction, causing solution to be slightly basic</p>
<p>EDIT: should be H2CO3 my bad</p>
<p>Question: should I even bother studying if I took a diagnostic MC test during my teacher’s AP chem review session (w/o studying of course) and got 52.5 raw points (~70%)? P.S. I only need a 3 to get my 8 credits but my college won’t let me test out of intro chemistry! Errr…oh well I guess I know which class I can catch up on my sleep!</p>
<p>Does anyone here have an example or worked out solution of a problem dealing with two ions in solution and adding something to precipitate one of the salts out (selective precipitation i think its called) and then figuring out how much of the salt precipitated before the other one started precipitating? <– crappy description i know but anyone know wat im saying? </p>
<p>forgot how to do those ill try to find an example</p>
<p>^ This may help
[Selective</a> precipitation of ions](<a href=“http://www.nyu.edu/classes/tuckerman/honors.chem/lectures/lecture_24/node4.html]Selective”>http://www.nyu.edu/classes/tuckerman/honors.chem/lectures/lecture_24/node4.html)</p>
<p>Are these on the exam??</p>
<p>So basically to summarize everyone’s answers to my ?, for these type of problems you need to:
-Check what the anion and cation of the salt make up in terms of acid and base strength
- Check what conjugate base it will yield</p>
<p>Is this the best way to do it?</p>
<p>Quick question: Does anybody know a good website that teaches rules for the equation writing part?</p>
<p>@bobtheboy: sadly yes lol, we also need to know how to find the amount in grams of precipitate and the percent precipitation and such : (</p>
<p>ANOTHER QUESTION</p>
<p>In order to produce SO3 the following two reactions take place:</p>
<p>4FeS2(s) + 11O2(g) -> 2Fe2O3(s) + 8SO2(g)
2SO2(g) + O2(g) -> 2SO3(g)</p>
<p>Calculate the mass of Iron IV sulfide required to produce 125 g of sulfur trioxide gas given that the percent yield for the SECOND reaction is 80%.</p>
<p>Im stumped, anyone know how to solve this?</p>
<p>I think the answer is about 117g.</p>
<p>first do 125 / .8 (=156 g)</p>
<p>Judging from the equations, work backwards with SO2.</p>
<p>156g SO3 x (1 mol/72 g) x (2 mol SO2/2 mol SO3) x (4 mol FeS2/8 mol SO2) x (~120g/1mol) = ~130 g FeS2</p>
<p>That’s what I would’ve done. Anyone wanna confirm?</p>
<p>@blstrs, everything else looks fine except for the fact you have to convert to moles before you take into account the 80% percent yield.</p>
<p>Wouldn’t it still be the same? (Order of Operations)</p>
<p>Well the way I did it was:</p>
<p>125g SO3 = 1.5625 mol
1.5625 mol SO3 / 0.8 = 1.953125 mol SO2
1.953125 mol SO2= 0.9765625 mol FeS2
0.9765625 mol FeS2 = 117.1875 g FeS2</p>
<p>Lol. blstrs, SO3= 80g per mol, not 72 ;)</p>
<p>it would be the same wouldnt it? since when u do it backwards, ud get the amount in grams then multiple by 80%</p>