If 36.2mL of 0.158 M CaCl2 sol is added to 37.5mL of 0.149 M Na2CO3, what mass of CaCO3 will be precipitated?
409421
n CaCl2= 0.0362*0.158=5.7196.10^(-3) (mol)\nn Na2CO3 = 0.0375*0.149=5.5875.10^-3 (mol)\nwe have   CaCl2+Na2CO3 = 2 NaCl + CaCO3 \nmol :      1  --> 1            --> 1  \n--> mol CaCO3 = min (n CaCl2 , n Na2CO3) = 5.5875.10^(-3) (mol)\n( that's means : 0.55875 g)
361205
1,000,000 millimoles= 1 mole \ntherefore u'd need 50/1,000,000 * 147kg of calcium chloride \nie. 0.00735kg and since 1000g=1kg u multiply this by 1000 and that gives u 7.35 grams.\nhope this helps... cheers
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
745889
We can solve this using a system of equations.\n\nLet x be the volume of the 25% solution. Let y be the volume of the 60% solution.\n\nThe sums of the volumes must total 140 ml:\n\nx + y = 140.\n\nThe mass of CuS04 in the desired solution must equal the amounts added through the use of both solutions.\n\n30 percent of 140ml gives 42 ml of "pure" copper sulfate solution.\n\nThe 25% solution would contribute .25x of "pure" copper sulfate solution. The 60% solution would contribute .60y of "pure" copper sulfate solution, so:\n\n0.25x + 0.60y = 42.\n\nHere is the system:\n\nx + y = 140\n0.25x + 0.60y = 42\n\nThere are many different ways to solve the system. I'll use substitution here:\n\nx + y = 140, so y = 140 - x\n\n0.25x + 0.6(140 - x) = 42\n0.25x + 84 - 0.6x = 42\n-0.35x = -42\nx = 120\ny = 20\n\nSince x is the amount of 25% solution and y is the amount of 60% solution, we need 120 ml of 25% solution and 20 ml of 60% solution.\n\nThis appears to match your answer!
57272
1 mole (also represented as mol)= 6.022x10^23.  This is just a number.  Molecules are represented in mol quantities because they're too small to work with otherwise.  So that when you are dealing with CaCl2 in these problems, you're actually using 6.022x10^23 molecules of CaCl2.  Additionally, when you see the molecular weight of a molecule, that is actually the weight of 1mol of that molecule.\n1M= 1mol/Litre.  You need 1.27M.  Divide 1.27 by 1,000 to find the number of mol CaCl2 per mL you need.  (1.27/1,000=0.00127mol CaCl2/mL)\nMultiply this by 147.5 to get the number of mol CaCl2 you will have.  (0.00127*147.5= 0.187325mol CaCl2)\nNow multiply this number by the molecular weight of CaCl2 to get your answer!  (0.187325*110.984= 20.79g CaCl2)
792910
This is not a complete equation. The products will be NaNO3 +  CaCO3\n\nBalance numbers:\n\n1,1,2,1
1064976
Q (heat energy) = m(mass in gram)*c(specific heat in J/gm/K)*t (change in temp)\nQ = 25 * 1 * (36-24)\nQ = 300 J
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
458346
2NaOH + CO2 = Na2CO3 + H2O
434401
I am pretty sure i did this right. using the CaCO3(s)as the limestone, i got 1178.934669 grams of CaO. (I didn't round the final answer, and w/in the problem itself, i rounded the atomic masses of Calcium (to 40.1), oxygen (to 15.9), and carbon (to 12)
706398
CaCl2 + 2AgNO3 --> Ca(NO3)2 + 2AgCl. AgCl is the precipitate.\n(80g cacl2) x (1mol cacl2 / molar mass cacl2) x (2mol agcl / 1mol cacl2) x (molar mas agcl / 1mol agcl)= g agcl= 206.6 g AgCl
880546
you should learn how to do your own homework hun you'll never be a chemist with someone else doing your work.\n\n\nuse the molar mass from the periodic table (units = g/1mol)\n\nCl = 35.5\nH = 1.00\n\nMolar mass, formula weight, molecular weight, formula mass = 36.5 grams HCl/1 mol HCl\n\n\nThis is a simple problem is you can't do this problem maybe you should try changing majors.  I hear history or english is nice.
117753
Okay you have a choice between Potassium oxalate (K2C2O4 ) and Sodium Iodide (NaI). \n\nPotassium oxalate is highly soluble (30g/100g) and sodium iodide is even more soluble(184 g/100 ml).\n\nWithout knowing the reactants it is almost impossible to predict a product, especially with two choices. Given those two products, I would expect it to be K2C2O4, but that is just a guess. What was in the solution?
808333
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
234325
what the previous writer has posted is true only for sodium. equivalence takes into account both the molecular or atomic weight of the species you're interested in and its valence - for example sodium has atomic weight of 23 and a valence of 1 (the ion is +1), hence the 23/1 conversion in the previous answer. This wouldn't be the same for, say, carbonate with mw ~60 and valence of 2 (= equivalence of 30)
877958
To solve this question, you have to find out the total number of moles contribute by individual solution, then divide by the total volume.\n\n\nEvery Na2SO4 molecule will dissolve and give 2 Na+ ion and 1 SO42- ion, while every KCl molecule gives 1 K+ and 1 Cl- ion.\n\n# of mole of Na+ = 0.2 * 0.018 *2 = 0.0072 moles\n[Na+] = 0.0072 / (0.018 + 0.015) = 0.218 M\n\n[SO42-] = 0.2 * 0.018 / 0.033 = 0.109 M\n\n[K+] = 0.15 * 0.015 / 0.033 = 0.068 M\n\n[Cl-] = 0.15 * 0.015 / 0.033 = 0.068 M
989197
204.54 g\n  Mass of 1.75 moles of Cl2 + mass of 3.5 moles of Na to react with that much Cl2 = 204.5g\n\nMW of Cl2 = 70.9\n\n70.9 x 1.75 = 124.08 g\n\n\nMW of Na = 22.99\n\n22.99 x 3.5 = 80.46\n\n80.46 + 124.08 = 204.54 g