dessskris Posted April 26, 2011 Report Share Posted April 26, 2011 3. As Kiwi said, because there are 2 ions Na+ and OH-4. 2 Reply Link to post Share on other sites More sharing options...
dessskris Posted April 26, 2011 Report Share Posted April 26, 2011 What is the structural formula for 2-methylbut-1-ene?I think it's:H5C2 - (CH3) C = CH2But apparently my textbook says it's:H2C = (CH3) CHHelp anybody? Thank you! Reply Link to post Share on other sites More sharing options...
Keel Posted April 26, 2011 Report Share Posted April 26, 2011 (edited) What is the structural formula for 2-methylbut-1-ene?I think it's:H5C2 - (CH3) C = CH2But apparently my textbook says it's:H2C = (CH3) CHHelp anybody? Thank you!It would be CH2=C(CH3)CHCH3You had one too many hydrogens on your chain, it would be C2H4... Edited April 26, 2011 by Keel Rubbish Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 26, 2011 Report Share Posted April 26, 2011 (edited) What is the structural formula for 2-methylbut-1-ene?I think it's:H5C2 - (CH3) C = CH2But apparently my textbook says it's:H2C = (CH3) CHHelp anybody? Thank you!It would be CH2=C(CH3)CHCH3You had one too many hydrogens on your chain, it would be C2H4...CH2=C(CH3)CH2CH3You were missing a hydrogen And desy...I don't know what your book is doing but a "but" and a methyl group is 5 carbons and that answer from your book has 3...and it actually makes no sense. Might want to check that you wrote that down right o.O Edited April 26, 2011 by Drake Glau 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted April 26, 2011 Report Share Posted April 26, 2011 (edited) does it look like this?drawing it on paper and scanning it would have been easier...but I'm too lazy to take out my scanner lolEDIT: Drake, which part is it that doesn't make sense?I believe we have the same textbook.. check out page 497 qn 3. Edited April 26, 2011 by dessskris Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 26, 2011 Report Share Posted April 26, 2011 My book is...too far away It doesn't make sense because you have a (CH3) connected to nothing and yes, that's what it would look like. Reply Link to post Share on other sites More sharing options...
dessskris Posted April 26, 2011 Report Share Posted April 26, 2011 lazy dudetextbook asks me to draw the mechanism for the reaction of 2-methylbut-1-ene with something. the answer key shows this as the alkene: Reply Link to post Share on other sites More sharing options...
iber2011 Posted April 26, 2011 Report Share Posted April 26, 2011 How do you guys solve equilibrium questions?? Is the general equation: x^2/(a-x)(b-x) for every equilibrium question?? Reply Link to post Share on other sites More sharing options...
Keel Posted April 26, 2011 Report Share Posted April 26, 2011 How do you guys solve equilibrium questions?? Is the general equation: x^2/(a-x)(b-x) for every equilibrium question??No. It really depends on the equilibrium equation as this determines the equation for Kc. Remember that 2A + 3B <=> C + 4D will give Kc as [C][D]4 / [A.]23 i.e. products over reactants e.g. the equation 2NO + Cl2 <=> 2NOCl has a Kc= [NOCl]2 / [NO]2[Cl]. I'm not sure why you have an x2 but to me that looks like the formula for Ka for a weak acid where the [H+] = [A-] therfore products = [H+]2. If you give us an example, maybe we could help better? 1 Reply Link to post Share on other sites More sharing options...
iber2011 Posted April 26, 2011 Report Share Posted April 26, 2011 I can't seem to be able to answer this question:At 1000C water decomposes according to the reaction below. if the equilibrium constant is 7.3x10^-18 and the initial concentration of water is 0.100M, find the equilibrium concentrations of hydrogen and oxygen. 2H2O -> 2H2 + O2. Please help me out in this. Thanks[H2]=5.3x10^-7[O2]=2.6x10^-7Don't know how to get that though Reply Link to post Share on other sites More sharing options...
Keel Posted April 26, 2011 Report Share Posted April 26, 2011 I can't seem to be able to answer this question:At 1000C water decomposes according to the reaction below. if the equilibrium constant is 7.3x10^-18 and the initial concentration of water is 0.100M, find the equilibrium concentrations of hydrogen and oxygen. 2H2O -> 2H2 + O2. Please help me out in this. Thanks[H2]=5.3x10^-7[O2]=2.6x10^-7Don't know how to get that thoughI guess since Kc is so small you could assume that the concentration of H2O remains constant which will just give you an equation of 7.3x10-18= 4x3/0.01 which gets rid of that nasty quadratic at the bottom. But the assumption only comes in with the dissociation of weak acid and bases, so when faced with something like this in an exam I'm not sure whether to accept the assumption or not. I guess you should accept it, its a chemistry question after all; shouldn't be testing your maths. 3 Reply Link to post Share on other sites More sharing options...
iber2011 Posted April 26, 2011 Report Share Posted April 26, 2011 OMG! That was soo simple!! haha! Thank you soo much and no you're not incompetent! Reply Link to post Share on other sites More sharing options...
Rigel Posted April 27, 2011 Report Share Posted April 27, 2011 I got this difficult question and i don't know how to work it out hehe.A student wished to determine the percentage by mass of calcium carbonate in an eggshell, she reacted 3.953g of the eggshell with 50cm3 of 1moldm-3 hydrochloric acid. After all the shell had reacted, the resulting solution was placed in a volumetrick flask and the volume made up to 100cm3. 10cm3 of this solution reacted exactly with 12.65cm3 of 0.1moldm-3 NaOH(aq). Calculate the percentage by mass of calcium carbonate in the eggshell. Reply Link to post Share on other sites More sharing options...
Keel Posted April 27, 2011 Report Share Posted April 27, 2011 I got this difficult question and i don't know how to work it out hehe.A student wished to determine the percentage by mass of calcium carbonate in an eggshell, she reacted 3.953g of the eggshell with 50cm3 of 1moldm-3 hydrochloric acid. After all the shell had reacted, the resulting solution was placed in a volumetrick flask and the volume made up to 100cm3. 10cm3 of this solution reacted exactly with 12.65cm3 of 0.1moldm-3 NaOH(aq). Calculate the percentage by mass of calcium carbonate in the eggshell.I didn't do this very well. You should realise at the beginning that this is a 'back-titration' question. Therefore it was 'wrong' of me to start with finding the moles of HCl first. I should have found the moles of NaOH first. Although I had to use the moles of HCl later, its just less confusing and more logical had I found nNaOH first. 2 Reply Link to post Share on other sites More sharing options...
Rigel Posted April 27, 2011 Report Share Posted April 27, 2011 (edited) The book says the answer is 94% And i didn't understand why you divided 0,03735 by 2, also, what does this number represent? Edited April 27, 2011 by Ipos Manger Reply Link to post Share on other sites More sharing options...
Keel Posted April 27, 2011 Report Share Posted April 27, 2011 The book says the answer is 94% And i didn't understand why you divided 0,03735 by 2, also, what does this number represent?Sorry my notation is poor. The line which states nHCl= 1/2 nCaCO3 = 0.018675 should actually say nCaCO3= 1/2 nHCl = 0.018675This is because if you think about the reaction equation (another thing is should have put) it is 2HCl + CaCO3 -> CaCl2 + H2O + CO2 Therefore the number of moles of CaCO3 must be half that of the number of moles of HCl.94% is double the number I obtained. So either they forgot to divide by 2 somewhere or I forgot to times by two. I've gone through my working one more time and I still stand by it. Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 28, 2011 Report Share Posted April 28, 2011 The book says the answer is 94% And i didn't understand why you divided 0,03735 by 2, also, what does this number represent?Sorry my notation is poor. The line which states nHCl= 1/2 nCaCO3 = 0.018675 should actually say nCaCO3= 1/2 nHCl = 0.018675This is because if you think about the reaction equation (another thing is should have put) it is 2HCl + CaCO3 -> CaCl2 + H2O + CO2 Therefore the number of moles of CaCO3 must be half that of the number of moles of HCl.94% is double the number I obtained. So either they forgot to divide by 2 somewhere or I forgot to times by two. I've gone through my working one more time and I still stand by it.Dilution factor of 2? Wouldn't that multiply your concentration (and moles consequently) and then that would double your percentage? I have no idea. We never practiced these at all and it makes me worried if one shows up on the exam... 1 Reply Link to post Share on other sites More sharing options...
Rigel Posted April 28, 2011 Report Share Posted April 28, 2011 Heh so what's the final answer? And you treat the 2 reactants in the 2 equations as if it were one right? And what is the dilution factor o.o Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 28, 2011 Report Share Posted April 28, 2011 The dilution factor is when you went from 50cm3 of stuff to 100cm3 of stuff. You doubled your volume. c=n/v to c=n/2v. You diluted it by halving the concentration which is diluting by a factor of 2. Reply Link to post Share on other sites More sharing options...
Rigel Posted April 28, 2011 Report Share Posted April 28, 2011 Thanks Drake, i always see you posting in here, are you a Chemistry God? Hehehe. Been teaching myself solutions, next class is on Friday, also on Sat, and i hope i can finish Cuantitative Chemistry and move on to Atomic Structure.BTWI'm following the IB Course Companion Book, is it a good book for revision/textbook? Reply Link to post Share on other sites More sharing options...
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