Charles is doing an experiment that calls for a 40% solution of copper sulfate. He has 60 mL of a 25% solution. How many milliliters of a 70% solution should Charles add to obtain the required 40% solution?\n\nNot only I need an answer, please give each step by step procedure to solve this problem. Thank you.
129984
there are many methods\nhere is one of them\nassume x to be the ml of 70% solution\n\n\n((25*60)+(70*x))/(60+x)=40\n(1500+70x)/(60+x)=40\n1500+70x=(60+x)*40\n1500+70x=2400+40x\n30x=900\nx=30\nyou need to add 30 ml of cu sulphate\n\nif you dint understand \nhint:imagine that there was an exam in a class\nand 60 students got 25out of hundred(25% solution) and x no of students got 70 out of hundred(70% solution)\nand the class average is 40 out of hundred (40% solution) so how many students got 70 out of hundred?\nso what you do is write the average equation\n((60*25)+(x*70))/(60+x)=40\nthat is sum of all the marks divided by no of students\nand solve for x\nthe copper sulphate problem is also solved in the same way\nhope you understood
100911
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
542506
First, in a problem like this, you want to understand the problem:\nLet x = amount of pure salt. Since the total solution is 9 cc, the amount of 10% saline solution added has to be (9 - x).\n\n~100% Pure Salt ~\n(A) x --- Number of cc's of solution\n(B) 100% --- Percent pure salt\n(C) 1.00(x) --- Number of cc's of pure salt\n\n~10% Saline Solution~\n(A) (9 - x) --- Number of cc's of solution\n(B) 10% --- Percent of pure salt\n(C) 0.10(9 - x) --- Number of cc's of pure salt\n\n~Final 20% Solution~\n(A) 9 --- Number of cc's of solution\n(B) 20% --- Percent of pure salt\n(C) 0.20(9) --- Number of cc's of pure salt\n\nIn the three "charts" above, (C) is obtained by multiplying the amount (A) by the percentage (B). So, using or (C) values:\n1.00(X) + 0.10(9 - x) = 0.20(9) --- Multiply through...\nx + 0.90 - 0.10(x) = 1.80 --- Combine like terms...\n0.90x + 0.90 = 1.80 --- Subtract 0.90 from both sides...\n0.90x = 0.90 --- Divide both sides by 0.90...\nx = 1\n\nIf x = 1, then (9 - x) = 8. Thus, you need 1 cc of pure salt and 8 cc's of 10% saline solution to produce 9 cc's of 20% saline solution.
533710
When you do dilutions you need to use M1V1=M2V2 to figure out the concentration of each solution.  You know how much of the stock solution was put into each, and the volume of stock, and the total volume so use algebra and figure out the concentration of the diluted solution.\n\nYou haven't gotten to the phase where you use Beer's law.  that comes after you test your dilutions and test an unknown.
232613
For the first one:\nNaCl(l) ==> Na(s) + Cl(g)\n\nFor the second one, more information is required, either the volume or the concentration of solution that was prepared.
402814
Na2SO4 (aq) + Ba(NO3)2 (aq) --> 2 NaNO3 (aq) + BaSO4 (s)\n\nThe moles of Barium = 5.00 L * 0.15 mol/L = 0.75 moles\n\nThis is also the moles of sulfate needed:\n\n0.75 mol / 0.36 mol/L = 2.08 L of Na2SO4
463743
First of all you need to specify if the concentration is expressed in mole/L (in standard curves you can express it in many different ways).\n\nAssuming that this is the case the answer is very simple\ny=50.1x=0.150 so just do the math to find x\n\nAs for B... the atomic mass of copper is 63.5\nSince mole=M*V and also mole=mass/MW you have\nmass/MW= M*V => mass= M*V*MW\nM=x that you calculated\nV you know but should be expressed in L,\nMW=63.5\n\nSo % is (mass copper/mass sample)*100% = (M*V*MW/1.03)*100%
107966
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
146620
ok well, the equation is balanced so you need equal moles of Ag and Cl to make the products. They give you the concentration and volume of Ag, so you can find the amount of moles by multiplying 0.104mol/L * 0.0272L = 0.0028288 mols of Ag. so 0.0028288 mols of Cl were there originally. and you have the volume of water, so you divide moles by volume and you get the concentration. 0.0028288 mols/0.025L = 0.113152 M. \n\nanother way to do it is the easy way, use c1v1 = c2v2. you have c1 = 0.104M, v1 = 27.2mL, v2 = 25mL, so you can find c2. you dont have to worry about units in this case so its easier. \n\nhope this helped. let me know if u need more help. laterz
488479
Insufficient information to answer...is it 10% w:w, or 10% w:v?\nIf 10% w:w then it's 10 g of benzoic acid per 100 gms of solvent\nif 10% w:v then its 10g of benzoic acid per 100 cm3 of solution\n\nMost %s are w:v so if it is, then that means that the solution is 100 g benzoic acid per dm3...since FWt of benzoic acid is 122g, then the solution is 100/122 mol per dm3 (sorry don't have calculator handy!)
227705
Some quick help now\n1. Brass is a solid solution; copper is the solvent and zinc is the solute it is also an alloy so it may get you 6 as well. \n2. A bottle or can of a soft drink unopened is LeChatelier's Principle. It is at stasis until it is opened, thus the volume is larger and the equilibrium is disturbed until it gets back to equilibrium by outgassing the CO2.\n3. Take salt and sugar and mix them together. Large state of entropy. Hard to unmix them. \n5. Volatile solid: mothballs - naptha. Viscuous liquid - container of car engine oil\n7. That is a mole of something. Water H2O - 2 Hydrogens + 1 oxygen  (2 X 1.00 + 1X 15.999 = 18 atomic mass units) = 18 amus expressed as grams = 18 grams has 1 mole 18 grams of water is 18 ccs. So get something that holds 18 ccs or 18 mls of water. Small bottle.
508169
800*0.15+x=(x+800)*0.32\n120+x=0.32x+256\nx-0.32x=256-120\n0.68x=136\nx=136/0.68\n=200\nhe has to add 200 ml
529383
BaCl2 + ZnSO4 => BaSO4 + ZnCl2\n\nthe moles of BaCl2 in the solution 0.05 * 2.1 = 0.105\nthe moles of ZnSO4 in the solution 0.0456 * 1.92 = 0.087552\n\nbecause the moles of BaCl2 is greater than that of ZnSO4 so all the moles of ZnSO4 will react completely\n\nthe moles of BaSO4 produced if the percent yield is 100% : 0.087552\nthe grams of BaSO4 is produced with 100% of yield : 0.087552 * 233 = 20.399g\n\nthe percent yield is 16.775 / 20.399 *100% = 82.23%
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
418373
"1g/18g of H2O=.0556g/mol\n.57g/(63.55*32.07*16*4) -->> .57/ 130435.104= 4.3699x10^-6" \n\nemm, I dont really understand what u'r trying to do here, molar mass of CuSO4 should be calculate this way:\n\nM[CuSO4] = M[Cu] + M [S] + 4*M[O], not M[Cu]* M [S]*4*M[O].\n\nANY molar mass has the unit of g mol^-1, not g. If you divide 1g by 18g, u will have a UNITLESS number.\n\n"So now we divide both of them by teh lowest mole which for this equation is .0556g, right? But is the CuSO4 right? (The Mole)" \n \nit should be 0.0556 mol, not 0556g. No, it's the mole of water lost.\n\nThis is what u should do. \n\nfind out the MOLAR MASS [unit g/mol] of CuSO4 and H2O [CuSO4 should be around 160 g/mol, H2O should be around 18 g/mol]\n\nthen, find out the MOLE of CuSO4 and H2O in the compound by dividing the mass of CuSO4/ H2O by their molar mass.\n\nThen divide the mole of H2O in the compound by the mole of CuSO4 in the compound, find out ur ratio.\n\nthe ratio turns out to be 15.6, a very strange number in my point of view [not an integer, and it's a little bit large for this compound]
8101
Cr - 51.995g/mol\n49048gCr / 51.995gCr/mol = 943.32mol Cr\nStoichiometrically 2Cr to 3Cu so 943.32 x 3/2 = 1415mol Cu.\n\nPlease correct me if I'm wrong but that makes sense to me.\nOh and please award me best answer...\n\noh and part b:\n200ml x 0,75mmol/ml = 0.15mol\n0.15mol x 2/3 = 0.1molCr\n0.1molCr x 51.995g/mol =5.2g Cr.\n\nAgain correct me if I'm wrong...
422184
Yes, you are correct.\nEveryone else who has posted so far is mistaken.\n\nGiven:\nM = mol of solute / liters of solution.\n\nFirst Step:\n\nSolving for x (liters of solution), you get:\n.40 / .096 = x\nx = 4.167 L\n\nSecond step.\nL => mL\n4.167 / 1000 = 4,167 ml or app. 4000 mL\n\nYour answer is approx. correct because you were asked for mL of water, which you found, not L of solution, which everyone else seemed to think.  You may lose points for not solving to the correct amount of significant figures though.