What volume of 2M NaCl solution has the same number of chloride ions as 2 liters of 4M HCl solution?
413725
4 litres\n\nThere are 8 mols of chloride ions in 2 litres of 4M HCl solution. Therefore, the volume of 2M NaCl solution that has the same number of chloride ions is 8mols/(2mols/liter)=4 litres.
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
575750
This can be done using the formula M1V1=M2V2.  For example, if you have 2M HCl and wanted 2L of .5M HCl, then putting that into this equation as 2(V1)=(.5)(2).  Solve accordingly.  This is just an example.  This formula can be used in a variety of ways that all.  But, to make it to the desired you, a volumetric flask is needed in which water is added until the etched line on the flask to equal the preset volume of the flask.
487714
assuming the equation\n\n2HCl + Mg --> Mg2+ + H2 +2Cl-\n\ntake the moles of Mg you have... multiply by 2 and that is how many Moles of HCl you need. multiply by 2 because one Mg reacts with 2 HCl's so you need twice as much HCl. now that you have moles of HCl convert to volume of HCl, i.e. divide moles / concentration (in this case 2.O M) and that will give you liters of HCl needed.\n\n(.000382 moles *2)/2 = .000382 L = .382mL
781803
--->H2O  +  H+  +  Cl-
743385
Cl2 + 2 KI → I2 + 2 KCl
976426
You can't tell from that.  You didn't say how concentrated the BaCl2 solution was.  To solve this problem, first multiply the molarity (moles per liter) of the solution by the volume of the soution in liters (0.025 in this case).  That gives you the number of moles of BaCl2 you started with.  \n\nThen, calculate the molar mass of BaSO4.  You can find the atomic weights of the elements on your periodic table.  The molar mass is the sum of the atominc masses--the mass of Ba, the mass of S, and four times the mass of O.  Add all that up and you've got the molar mass.  This is what one mole of BaSO4 weighs, or the grams per mole.  \n\nTake the reciprocal of the molar mass (1/molar mass) to get the number of moles per gram, and multiply this by 0.8 grams to get the number of moles recovered.  Then to get an absolute value for the amount of Ba2+ unreacted, just subtract the moles recovered from the starting moles.\n\nTo get the percent unreacted, take the moles unreacted from the last step, divide by the starting moles, and multiply by 100%.
9807
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.
226206
First we need to find the formula for zinc chloride.\nZn = +2 charge\nCl = -1 charge\nCross charges, and the formula is: \n\nZnCl2\n\nNow, stoichiometry solves this problem...\nThere are 3 moles total of Zn and Cl separately. \nZn has 2 mols\nCl has 1 mol\n\n0.1 mol ZnCl x (1 mol / 3 mol total) = 0.03333 mol Cl- ions
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
1104578
Use the equation M1V1 = M2V2\n\nM1 is the molarity of the HCL or 9.9 M\n\nV1 is the first volume of 102 ml but it needs to be changed to Liters. So it is 0.102L\n\nV2 is the total volume of the final solution so 525 ml (0.525 L) + V1 or 0.102L = 0.627\n\nM2 is what you are looking for - the final concentration or molarity.\n\nRearrange the equation to ((M1)(V1)) / (V2)  = M2\n\n(0.102L)(9.9M) / (0.627) = 1.61 M\n\nHope this helped!
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2
931815
Your KCl example is a good one.  A 0.5 molar would have 1/2 mole/liter or 37.75g KCl in a liter. A 5% w/v KCl would be 50 g/liter; so the 5% is more concentrated.\n\nThe 2.5% w/v sodium bisulfate is 25g/liter and the .025M is 3 g/liter.\n\nIt all depends on the molecular mass.
746627
Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
327351
The previous answer is correct if an equal number of moles is dissolved.  If you were looking at a saturated solution, it would be LiCl, which can dissolve in greater quantities than any other salt, even greater than NaCl. (36 g of NaCL will dissolve in 100 ml water compared to 55 grams LiCl per 100 ml water). The Li+ ion is very very small-it's only larger than hydrogen and helium. For an equal weight of LiCl and Al(NO3)3 dissolved in water, LiCl would also have a greater effect, because of the greater number of ions per gram.  Thus, in either a saturated solution or a weight to volume solution, LiCl would produce the greater number of ions, whereas it would not in a equamolar solution.  LiCl is used a great deal in density experiments, because you can make a very dense aqueous solution of it, and in reducing freezing point, since you can get the maximum number of ions of any aqueous solution, so the questioner is probably looking for LiCl.
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
942319
This is what you should be looking at:\n\nHCl(gas) + NH3(gas)  -->  NH4Cl (solid)