You are given the value of the distance between Na Cl: it is 2.80 x 10^-8 mL \n\nPLEASE SOME CHEMIST!!!! HELP MEEEEEEE
133053
why dont you just ask avogrado for there number.  if they see you are a nice person, just talk to them.  let them know you want to get to know them better and would like there number.
422670
You have a common ion effect problem\n\nPbCl2 <=> Pb+2 +2Cl-\n\nKsp=[Pb+2][Cl-]^2\nYou know that [Pb+2]=5.5*10^-3\nAlso from the reaction you know that the amount of Cl- coming from the PbCl2 is double the amount of Pb+2.\nSo if C is the concentration of Cl- coming from HCl, the amount of Cl- at equilibrium is [Cl-]=C+2*[Pb+2] \n\nSo Ksp= [Pb+2]* (C+2*[Pb+2])^2=>\nif we assume that C>>2[Pb+2] then C+2*[Pb+2] =C and \nKsp=[Pb+2]C^2 =>1.7*10^-5= 5.5*10^-3 C^2 =>\nC=55.6 * 10^-3 but 2*[Pb+2]=11*10^-3, so we can't do the approximation...\n\nLet's name [Pb+2]=p for simplicity\n\nKsp=p(C+2p)^2 =>\nKsp/p= C^2+4pC+4p^2 =>\nC^2+4pC+4p^2-Ksp/p=0 =>\nC^2+2.2*10^-3 C -2.97*10^-3 =0\n\nC=0.5*(-2.2*10^-3 + SQRT( (2.2*10^-3)^2 +4*2.97*10^-3) ) =>\nC= 5.34*10^-2\n\nIf you added Va L of HCl, then\nC=mole HCl/Vtotal = Ca*Va/(Va+0.12) =>\nCVa+0.12C=CaVa =>\nVa=0.12C/(Ca-C)= 0.12*0.0534/(0.12-0.0534) = 0.0962L or 96.2 ml
21443
S8 + 4 Cl2  ---> S2Cl2\nHence here 1 atom of S reacts with 1 atom of CL\n\n21.51 gm S8 contains 0.67 moles of S atoms and .67/8 moles of S8\n\n82.98 g Cl2 has 1.16 moles of Cl2\n\nHence since 2 atoms of S make 1 S2Cl2 atoms, we get\n\n0.67 / 2 \n= .335 moles of S2Cl2\nmass= 135 x .335\nmass= 45.225 gms of S2Cl2
462746
Going from 10^12 cells/mL to 10^10 is a 1:100 dilution.\nIf you want a final of 2mL, you need to add 20uL of your stock solution.\nSo do 20uL of stock (10^12)+ 1980uL of salt solution = 2000uL (10^10), which is the same as 2mL.
474098
NaCl + AgNO3 -> NaNO3 + AgCl\nx                                             x\nKCl + AgNO3 -> KNO3 + AgCl\ny                                         y\nlet x be the moles of NaCl and y be the moles of KCl\nwe have\n58.5x + 74.5y = 4\n143.5 (x+y) = 8.5904\n\n=> x = 0.028 y=0.031\n\nthe mass percent of  each substance\nNaCl = 58.5*0.028 / 4 *100 = 43 %\nKCl = 74.5 * 0.031 / 4 *100 = 57 %
489988
the moles of NaCl 0.416 * 0.35 = 0.1456 moles\nthe grams of NaCl 58.5*0.1456 = 8.5176g\n\nyou got the right answer
287622
if its 2% by mass, thats 2 grams of KCl for every 100 grams of solution its in. \n\n2g KCl= 2g / (74.55g/mol)KCl = 0.0268mol KCl\n\nBecause there is one mole of Cl in KCl, this is also the number of moles of Cl in 100 grams of solution.\n\n0.0268mol Cl * (35.45g/mol)Cl = 0.951gCl/100gsolution\n\nThis is 0.951%\n\nnote: 35.45g/mol is the molar mass of Cl and 74.55g/mol is the molar mass of KCl. These were obtained from the periodic table.
39691
By a particle diagram, do you mean a graph? If you do, then draw a graph with an X (time) and Y (concentration) axis. When you put NaCl into water, the concentration of NaCl will be 100%. But as time goes by, the NaCl molecules will be pulled away from each other by polar water molecules (NaCl concentration will decrease), causing the appearance of a Na concentration and a Cl concentration. These two concentrations will increase from zero. I believe the slope of the two latter elements will be (theoretically) the same. So basically you should have two lines (increasing NaCl line and one line (which actually includes the 2 separate lines for Na and Cl because they are the same) for Na and Cl crossing each other like an X.
553750
1 litre (1000 ml) of 1 M solution of NaCL contains 23 + 35.5 gms =58.5 gms (ie 1 molecular wt of sodium chloride in gms dissolved in 1 lit of solution)\n\nHence 1 lit of 0.360 M solution  will contain 0.360 X 58.5 gms or 21.06 NaCl (Sodium Chloride) \n\nTo find out exact vole in ml which contain 21.14 find volume by this formula 2.14 divide by 21.06 multiply 1000 , (ie 2.14/21.06 * 1000 ) 101.6 ml\n\nPS\nbasic assumption is 0.360 M indicates .360 Molar solution. \n(ie: M stands for Molar)
622656
Na2CO3 + 2HCl-> NaCl + H2O + CO2
675309
PV=m/MM * RT\nPV/RT = m/MM\nMM = mRT/PV = 0.800g * 0.08206 * 373/ ((750/760) * 0.256) = 96.9 g/mol\n\nCHCl is 47.5 g/mol, so molecular formula is C2H2CL2.
381945
Na is in column 1 with 1 electron in the outer ring. Cl is in column 7(17) with 7 electrons in the outer ring. Each  wants to havae 8. So Na gives 1 away to Cl and both are happy. Cl does not want 2. then the Na+ and Cl- are attracted to each other. \n\nCu is a metal. it lies to give away electrons so it is conductive. CuCl2 is a compound where Cu has already given 2 electrons to the 2 Cls so noes no want to further shift electrons so does not conduct. If you put it in water where the Cu++ and the 2 Cl- can float around, they will  conduct through the ions.
351029
Suppose that we have r milliliters of the 8% solution.  This means the amount of salt in it is .08r.\nSuppose we have s milliliters of the 3% solution.  This means that the amount of salt in it is .03s. \nWhen we mix them together we have r + s milliliters total, and the amount of salt in it is .08r + .03s.  If the resulting mixture is a 5% solution then this means that  \n(.08r + .03s)/(r + s) = .05.\nSo .08r + .03s = .05(r + s) = .05r + .05s.  \nThen .03r = .02s, or r/s = .02/.03 = 2/3. \nBut we know that s = 600 mL, so r = (2/3)s = (2/3)600 = 400 mL\n\nAdd 400 mL of the 8% solution to 600 mL of the 3% solution and get 1000 mL of 5% solution.
251640
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
3581
Well lots of chemistry questions lately....finals coming up? :)  Good Luck!!!  \n\nTo calculate this type of question first of all calculate the #of moles of HCl used.  So \nyou used 2100 mL = 2.1 L HCl \nThis HCl is 0.3 mol/L so you used:\n\n2.1 L HCl* (0.3 mol HCl/L) = 0.63 mol HCl\n\nnow use the ratio of Cl/HCl  (which is 1 to 1)\n\n0.63 mol HCL *(1mol Cl/1mol HCl)= 0.63 mol Cl ion\n\nSo far so good I hope.  \n\nNow do your solution diltion to calculate volume,\n2.1L + .2385L = 2.3385 L of mixed solution\n\nNow simply divide \n0.63 mol Cl/2.3385 L solution = 0.269 mol Cl ion in final solution.\n\nAgain these numbers are quick and dirty but the method of solving this type of problem is the same.
368037
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
318012
NaOH + HCl --> NaCl + H20\n\nn(NaOH) = 1.00 x 32.6cm3\n        = 1.00 x 32.6 x 10^-6\n        = 3.26 x 10^-5  mol\n\nn(HCl) = 1.00 x 25cm3\n       = 1.00 x 25 x 10^-6\n       = 2.5 x 10^-5  mol\n\ntherefore, NaOH is in excess.. hence the final solution is alkaline, which means the pH is more than 7.\n\nDue to the limitation of my knowledge, i can't give you the exact pH value. Just hope that it will help you to continue solving...