tell me about laboratory balances?
928755
Most modern lab balances are digital. They come in a variety of types, from those that measure with a precision of 1gram to those that can measure to .0001 grams. The more precise the instrument, the more costly.  I have a number of .01 gram scales that cost about $400 US and 1 .0001 gram scale that cost $2000 US
929685
When you have weak electrolytes, (equillibria in general) you should make the following table.\n\n1. write the reaction\n\nHNO2 <=> H+ + NO2-\n\n2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\n\n3.Assume that an unknown amount "x" of the reactants react.\nThen write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+  and NO2-, so x mole/L HNO2 will  give x mole/L of each and our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\n\n4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is \n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. .. .. x .. .. .x\n\nKa= [H+][NO2-]/[HNO2] = x^2/(0.5-x)\nNow you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.\nThen Ka=x^2/0.5=4*10^-14 =>\nx=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).\n\nFor a simple level of chemistry we would now say\npH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85\n\nHowever for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH. \nThere are 2 approaches for this case, here is one of them.\nRemember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5 .. .. .. 10^-7\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. 10^-7+x. .x\n\nand Ka=(10^-7+x)x/(0.5-x)\nNow we have to solve the quadratic so we can't do any simplification\n\n0.5Ka-Kax=10^-7x +x^2 =>\nx^2+ (10^-7+Ka)x-0.5Ka=0\n\nThe only acceptable solution is the positive one\nx=0.5(- (10^-7+Ka)+ squareroot (D))\nwhere D=(10^-7+Ka)^2-4*(-0.5Ka)=\n=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)\n=9*10^-14\n\nand x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7\n\nso now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70
986618
The equation is already balanced. Count the Ag's on each side = 1. Count the Na's on each side = 1. Count the Cl's on each side = 1. Count the NO3's on each side = 1. One can count the N's on each side = 1, or the O's on each side = 3.\n\nBut one could also consider the question another way:\nAg+ + NO3- + Na+ + Cl- ===> AgCl + Na+ + NO3-
336505
Just post 'em on Yahoo! Answers and someone will balance them for you!
382713
The best method is pH meter (other than color indicator). \nTo measur the exact end point, what you have to is:\n\nStart the titration\nInsert a pH probe inside the solution\nNow start adding base to the acid and swirl the solution properly\nKeep track of your pH readings..\nYou will see a sudden jump (some thing like from 5.6 to 9.0 even with one drop addition)\nthats end point.
468477
The approximate values are as follows:\nHbA1c....BSL(mg/dl)\n6%.........120\n7%.........150\n8%.........180\n9%.........210 \n\nand so on...But these values (ie, the normal range for HbA1c) maybe differ from lab to lab or also based on the technique used to determine the HbA1c level.
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
894918
It's a convenient way to count atoms or molecules in amounts that can be measured with a balance in the lab.
1146046
Here are some Lab Apparatus \nhttp://library.advanced.org/11430/research/apparatus.htm\nhttp://www.charltonsci.co.uk/charlton_scientific_015.htm\n\nhere are all the Lab Apparatus (click on them to get info one them)\nhttp://en.wikipedia.org/wiki/Category:Laboratory_equipment
239993
Links at\n\nhttp://www.itt.edu/staff/boyter/links/index.html#CHEMISTRY_CHROMATOGRAPHY
851400
the first equ is balanced, b/c on the left side u have Pb-1, C-1 O-3 on the right side u have Pb-1 C-1 O-3 so since both side are equal it is balanced\n\nans for question 2 BeCl2 + 3O2------Be(ClO3)2\nBe-1                                  Be-1\nCl-2                                    Cl-2\nO-was 2 now 6                                     O-6
1038474
http://www.chem.purdue.edu/gchelp/howtosolveit/Solutions/concentrations.html
78569
try http://www.wfu.edu/~ylwong/balanceeq/balanceq.html which is an interactive site OR try http://members.aol.com/profchm/balance.html if you want a lot of examples.
1057804
try this\n\nhttp://en.wikipedia.org/wiki/Category:Laboratory_equipment
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
662629
Wrrite the balanced eqution.  The acid and base will react to give water and the salt.  There will be so much acid left.  Calculate the pH of that amount of acid with the salt formed present.  THis is a buffer calculation.
974789
I'm not positive, but I think a platform balance is weighing device with a flat plate mounted above a balanced beam.  A triple beam balance works by sliding the bars until you get the scale level (like the scale at a doctor's office)\n\nIt depends on the scale, but usually a platform balance will weigh up to 0.1 of a gram while the triple beam balance will read up to 0.01 of a gram.\n\nIt's been a while since I had a lab, but I hope that helps.