I posted earlier but forgot some data, so here I go again WITH the missing data :D\n\nI'm just going to admit it...I don't get this...and my teacher has a tendency to do labs whenever I'm absent and then he tells me to figure them out on my own *sighs*.\n\nAnyhow, if anyone could help me with these Analysis questions for a chem lab that would be great.\n\nJust click on the link and it will take you to a picture of the lab that I scanned in. I also included the data I found on the picture that are needed to complete the questions. There are 5 Analysis questions and I need to complete all 5 of them...just answers won't really help, I actually need to know how you got there...so please show as much work as possible...I know it's hard on the computer.\n\nhttp://smg.photobucket.com/albums/v695/moonlight1811/Homework/?action=view&current=lab.jpg\n\nThe min. temp is 7.96C and the max is 24.10C.\n\nIf anyone could help me with any of the 5 questions I would be very grateful. Thank you :D
130905
This is going to sound mean, but it sounds like you're missing lessons a fair bit. Try and attend them then you will get the benefit of hands on experience and the teacher's guidance.\nIf there is a very good reason why you missed one, talk to him and he might be willing to help.
411718
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
529024
I'm not going to solve these for you.  My homework days are far behind me.  I will try to teach you how to solve them, though.\n\n1.  Keep your eye on the moles and don't get confused by the changing volumes.  The final solution is 0.00436 M  and you know the final volume so calculate how much KNO3 is there (=0.00436 M x Vf, in liters).  That will tell you how much was in 25 mL of the second solution.\nSo how much was in the total second solution?  Multiply the moles above by five (25 mL goes into 125 mL five times.)  This is also how many moles was in original 20 mL solution.\nSo, to find the molarity of the original solution, divide the number of moles by the volume, in liters (20 mL = 0.02 L).\n\n2.  Same thing goes for this one.  Multiply 1.66 M by .037 L (remember that M is in terms of liters, not millilieters!) to get the moles in the first solution.  Do the same for the second.  Add the two volumes together to get the resulting volume (in liters!!!)  Divide the sum of the moles in both solutions by the total combined volume and you have your answer.\n\n3.  Same would go for this one if you hadn't left out what volume of solution the Pb++ was in.  Anyway, the way to do this one is to notice that the reaction of Pb+2 ion and SO4-2 ion is a one-to-one proposition.  So, if you figure out how many moles of Na2SO4 you have in 0.009 grams of the stuff, you will also calculate how many moles of Pb(NO3)2 you have.  Divide that by the volume in liters and you have your answer.  [Nice touch by Teach to throw in univalent species to cross you all up.  These are irrelevant to the reaction you're considering.  Still, I like that.  Makes you think.]\n\nGood luck.
384026
Look this one up on the internet.  I doubt you're gonna find anyone here to do it for you.  That's too much information.  Don't you have a textbook? Use that.
489202
You might want to try to re-read the questions.  You don't appear to be understanding what is written there.  (I call it a brain fart, but three out of 28 indicates a little lack of understanding)\n\nIf you have re-read the problems, then I offer the below:\n\nYou might want to go over your notes about what a mole is.  It is an amount and a number.  It is a fundamental conceptual building block for the study of chemistry.\n\n  It looks like you expect to be able to plug and chug on these.  Which you can, once you start thinking in terms of atoms and molecules.  Plugging and chugging BEFORE you understand what these little pieces are doing is a recipe for a C.\nIt IS the ATOMIC theory of matter, don't you know.\n\nSo, once you've absorbed that, re-read these three problems.\n\nStill no joy?  Read on, Gunga DIn\n\nFirst H(g) is NOT the element's standard state! I thought everybody learned in grade school about H2 gas - the simplest molecule?  Iodine, I leave to you  (hint a gas???!!! not)\n\nSecond lets see you're making umpteen moles of something and calculate the dS.  that was not the question.  (hint: something about ONE mole)\n\nThird read the problem. your chemical equation is from I don't know where but not from what is in the problem. (hint: products on the LEFT???)
531456
Here is a great site, made specially for teachers, that has many tutorials, plus links to others\n\nGood luck\nhttp://sciencepage.org/chem.htm
319335
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
410491
well theres alot of material i can provide you with but i dont know what particular experiment you are designing as there are always different experiments within an experiment.\nanyhow, i think this is an example of an experiment for photosynthesis and respiration and their pH levels.\n\nhttp://www.science.siu.edu/plant-biology/plb337/old_st_papers/emmert.pdf \n\n\nif you are doing the same experiment, i guess you will find this one really helpful.\n\n.. Good Luck
379891
First, we have to check if the equation is balanced, which in this case it is. So, we can continue.\n\nNow, let's find out the total number of moles of Cu and HNO3.\n\n# mole of Cu = 4.55 / 63.55 = 0.0716 mole\n# mole of HNO3 = 6.29 / (1 + 14 + 16*3) = 0.0998 mole\n\nWhat we have to do now is to determine which compound (Cu or HNO3) is the limiting reagent. One easy way to find out is to assume either one compound will be completely used up and use the stoichiometry coefficient (the numbers in front of the compounds) to determine how many moles of the other compound is required. \n\nFor example, assume all the Cu will be reacted, the number of moles of HNO3 needed = (0.0716 / 3)*8 = 0.191 moles\n\nSince it requires more HNO3 than we have, HNO3 is the limiting reagent. Now, apply the 46.4% reaction yield to this question.\n\nFor every 8 moles of HNO3 that react, it produces 3 moles of Cu(NO3)2, since we have 0.0998 moles, IF for 100% reaction yield, we will produce\n\n(0.0998/8)*3 = 0.0374 moles of Cu(NO3)2\n\nSINCE the reaction is just 46.4%, ACTUAL yield will be\n\n0.0374*0.464 = 0.0174 moles\n\nMass of Cu(NO3)2 = 0.0174 * (63.55 + 14*2 + 16*6)\n= 3.26 g
59130
Hey I'm studying physical and chemical equilibrium too, currently.The answer is that it will definitely have units when the concentrations have different powers.Just try and work it out with the units and it will be very clear.
449412
how were you measuring the pH? if it was a pH meter maybe it wasn't calibrated correctly/ recently?? or if it was with litmus paper.... maybe it went bad? I'm not sure if it can... check and see if there could have been an expiration date. i don't think temperature would have that drastic of an effect unless you did the experiment at pH7 and a seriously high (like about 60C) temp and the other ones at room temp or 37C (body temp)... sorry i cant help anymore then that, that's very strange!
206344
check the valence, balance the equation, look up the atomic weights on that EVIL periodic chart, make sure you understand WTF a "mole" means, and...aw hail, who'm I tryna kid, I don't fukn reMEMber!!!!\n\np.s. Mr. Lewis, despite your arrogance looks like you SUKKED as a teacher! †
351460
I am not a chemistry student but I make this research for you and maybe it helps you http://www.chem.latech.edu/~deddy/chem104/104Antacid.htm\nhttp://www.chemcool.com/regents/acidsbasessalts/aim3.htm\n\njust copy those links and you should be on your way to solve ur problem.
116399
see here http://www.people.vcu.edu/~mikuleck/courses/ch4/sld029.htm
548445
maybe what your teacher is trying to say here is the body temperature. now search for that answer. it's your homework.
445789
This compound is an amine.  You can identify it by the NH2 (amine group) at the end.  You did a fair job with the structure considering the limits of the text here.  I am sure you realize that two of your carbons do not have the correct number of bonds.
446019
sorry, if my mother was still alive - chemist/physics major, then she could have helped... try this site:\n\nhttp://chemistry.about.com/