Im helping my brother with his homework and I am completely stumped on how to set up the problem. I dont want the answer just how to set the problem up please!
213284
To do this problem we need to do some convertions.  Convertion factors are simply ratios that allow us to change the units of a certial quantity.  For example: the molar concentration (molarity) is a convertion factor that is a ratio of the moles of dissolved substance to the volume of solution (moles/L). Using the concentration  we can convbert moles to liters or liters to moles depending on what we have to start with.  Another convertion factor that we will need is the molar mass of KOH.  This is the ratio of the mass of KOH to the moles of KOH (this is usually measeured in grams per mole (g/mol)).  We can find this number by looking up the molar masses of K, O, and H on a periodic table and adding them together. K =39.0983g/mol, O = 15.9994g/mol, and H = 1.00794g/mol. \nAdding these together we have a molar mass of 56.10564g/mol for KOH.\nSchematicaly this is the road map we will follow to find our answer:\n\n-------(moles/L)-------(g/mol)\nvolume <------> mole <-------> mass\n\nWhere volume, moles, and mass are actuall ammounts and the convertion factors between them are on the arrows.\n\nNow for your specific problem we will start with the given volume 750mL.  First we will convert it from mL to L then convert it to moles of KOH then finally to grams of KOH.\n\n750ml(L/1000mL)(15.0mol KOH/L)(56.1056g KOH/mol KOH)= 631.188g KOH\n\nTo three significant figures our answer is 631g KOH.  This is a big number but a 15.0 M solution is very concentrated so it is not surprizing.
89131
They used log tables.\nSimple problems like this could also be solved by using Newton's method.
321442
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
604333
Let the numbers be n and n+1 (consecutive)\n\nthen the answer is 2n(n+1)\n\nYou know, I can tell you're really trying, (not just fishing for homework answers) because you didnt ask the whole question, just the part you needed help with setting up properly.\nGood Luck!
118902
Oh, come on. I answered your first one, and now you flood the board with your homework? Are you really that lazy to think it's easier to post your homework here? These problems are all the same, with different numbers. Do it yourself, I already showed you how.
251639
this is a little more than Y answers can handle...you're better off looking at a tutorial online and using your book....good luck...you'll get it. C is not that hard to learn
395556
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
468582
there is no problem to be solved, you already have one variable all by itself on one side. what your book is probably asking is for you to graph this. the graph should be one straight line going up and down from the -3 spot on the x-axis.
453548
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
192286
Its tricky work, i need digrams to explain to you so your better just following this link, I think it will help you out.
373414
So lets roll...\n\nmathematically, set it up as follows:\n\nLet x = the number of kg of water @ 82degC that must be added\n\nEnergy = m * c * T\n(m = mass, c = specific heat of water, T = tempearature)\n\nNeglecting convective and radiation heat losses:\n\n (x kg)*c*(82 C) + (0.2kg)*c*(14 C) = (x + 0.2 kg) *c*(36 C)\n\ndividing by c and re-arranging:\n\n82x + 2.8 = 36x + 7.2\n\nx = 0.096 kg
535136
all you need to do is go to www.webmath.com, it'll help u there!\n\nI always do my math problems to that site if ever i get stuck on something! It's helpful!
229478
The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
445079
Well hey, you need help, ask the teacher! I know how to do this, with help from my nerdy Calculus BC friend, but it's really not that hard once you think about it...
460677
I see your preferred method is ask it here and have someone else figure it out for you...\n\nI think you'll learn more if you try them yourself.\n\nFor the addition and subtraction I would write them as numbers on top of each other and use carrying and borrowing rules:\n\n176\n+43\n-----\n\n.492\n-273\n------\n\nFor the multiplication, you would again put them on top of each other and multiply a digit at a time, then add the results.\n\n136\nx21\n------\n..136\n272\n-------\n\nFor the division, I would suggest long division\n.. _____\n3 ) 429
457468
Word problems like this can be hard.  Everybody has problems with them sometimes.\n\nIt does help to draw it out.  So you draw a little rectangle like this:\n ______________\n|                            |\n|                            |\n|______________|\n\n2 of the sides are 120 m and the other 2 are 150 m.  So you take 2 times each of these numbers and add them together:\n2(120) + 2(150)\n\nThat's your perimeter, in meters.  Now if your poles are 2 m apart, you don't need one for every meter you get in your answer, you only need 1 every 2 meters.  So you divide your answer by 2, and that gets your number of poles you need to put around the field.
41517
Oh you naughty boy...I AM going to tell your teacher. If you don't learn to do your own work you will never amount to anything. \n\nYahoo!Answers is about helping with simple questions...Not for you to submit your homework online!