"what is the chloride concentration of 2% KCl?
603847
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.
558772
the correct answer is 1.30.  the answer would be two if the concentration was 0.01M.  you just take the negative log of the concentration of hydronium ion.
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
867786
Let's take potassium representing group 1.  All the others will do the same thing.  Here is the reaction:\n\n2K(s) + 2HCl(aq) --> 2KCl(aq) + H2(g)\n\nThe reaction for group 2 will be similar.  Let's take calcium  representing group 2:\n\nCa(s) + 2HCl(aq) → CaCl2(aq) + H2(g)\n\nHope this helps.
796075
At 10°C, saturated solution of potassium chloride is more concentrated than potassium nitrate. At this temperature, potassium chloride completely dissolves while potassium nitrate would form some precipitates. At 60°C, it would be the same since both would completely dissolve at this temperature.
969868
What parameters do you already have?  I guess you have an a/c weight? The CL equation is a standard with CL= L/(1/2 density*V^2*S)  where S is the Area.
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
95222
Potassium Chloride (KCl) has a solubility of .4 g/mL in Ethanol (at room 25 C I would assume), yes, it does dissolve, but barely.
152231
I believe it's sodium chloride
489184
I think they're both 1.ooM.  \n\n          HCL -> H+ + Cl-\n         1mole   1mole  1mole\nUsing mole ration, the hydronium and chloride ions will both be 1 M concentrations. :)\n\nI think that's right.
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
945429
Equation becomes CuCl2 + 2KI --> CuI2 + 2KCI\n\nProducts are Cuprous Iodide and Potassium Chloride
969279
You must be talking about Potassium Chloride solution and not Potassium Chloride in molten form. So it would contain K+, Cl-, H+, OH- ions, and once electricity is passed through, K+ and H+ migrate to the cathode. Hydrogen being lower in the electrochemical series get discharged first as Hydrogen gas, (2H+ + 2e- ----> H2) and both Cl- and OH- ions migrate to anode but because of concentration effect, Cl- loses electrons and get discharged as Cl2 gas. (2Cl- - 2e- ---> Cl2) Thus K+ and OH- ions are left behind. These form Potassium Hydroxide and all group I hydroxides are alkali hence the resulting solution is alkaline.
118416
MY ANSWER IS:\n\n1 mol/g of LiCl = 35.4g
78988
I2 + 2NaCl ---> 2NaI + Cl2\n\nSodium Iodide and Chlorine
307315
=  because 2kL is 2 kiloliters and 2 kiloliters is 2000L or liters
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%.