how do i calculate the molartiy of 200g of KCL in 800ml of water?
404275
divide 200g by the molecular weight of KCL and then divide that by .8
414222
For each, you have to calculate the number of moles needed from the volume and the final molarity:\n\nFor potassium iodine:\nV= 400ml= 0.4L\nM= 0.2M= 0.2 moles/L\n\n#moles KI required = 0.4L * 0.2moles/L= 0.08moles KI\n\nYou can then calculate the mass of KI needed by using the #moles calculated above and the molecular weight of KI\n\nMW of KI= 166 g/mole\n\n# grams KI required = 0.08 moles KI required * 166g/mole\n# grams KI required = 13.28g KI required\n\nYou calculate the answer for FeCl3 in the same manner.  I think you can handle it from here.
812576
From the reaction you had heat produced\n\nQ=m*Cp*ΔT\n\nfor water Cp= 1 calorie/(gram*Celsius)\n\nYou have a total of 50+50 =100mL and since d=1.00 g/mL you have 100 g\n\nso Q =100*1*(31.43-25.32) = 611 cal\n\nyou mixed mole HCl= C*V= 1*0.05= 0.05 with\nmole NaOH =C*V =1.1*0.05 =0.055 which means that, since the stoichiometry of the reaction is 1:1, HCl is the limiting reagent\n\n0.05 mole HCl give 611 cal\n1 mole HCl gives x\nx=611/0.05= 12220 cal/mole =12.22 Kcal/mole
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.
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.
780066
V = nRT/P\nV = (25g/44g/mol)(0.08206Latm/molK)(298K)/1atm\nV = 138.9 L
755177
1. Make sure that HCl is standardized.\n2. Write the balance equation:\n\nCa(OH)2 + 2HCl ---> CaCl2 + 2H2O\n\n- this means that 1 mole of Ca(OH)2 reacts for every 2 moles of HCl\n\n3. Get the number of moles of HCl reacted by multiplying the volume used to react Ca(OH)2 by the actual concentration of HCl (value from standardization). Divide the result by 2 - and this is now your moles Ca(OH)2.\n\n4. Get the molar solubility of Ca(OH)2 by dividing its number of moles by its volume. Unit is in gram per liter.
318519
One mole of NaOH weighs 40gms. 200mmol is 0.2 of that and 100ml is 0.1 of a litre. \nSo you would need 40 X 0.2 X 0.1gms of NaOH.=0.8gms\n\nIn reality you would likely dissolve 8gms of NaOH into a 1L volumetric flask and fill it to the mark. You would then measure out 100 mls of this solution. Weighing out 0.8 gms of NaOH is inaccurate because it reacts with CO2 in the air and absorbs water vapour.
1006885
I think you are leaving out the first step.\n\nRemember that 12.0 M means 12 moles of HCL per liter of aqueous solution.\n\nTo understand the equations so that you can use them correctly, first think that you will have to add some amount of water to some amount of the concentrated solution to make it the dilute solution.\n\nso, 12 mol/liter , assuming you had a liter of the solution, you keep the same number of moles and add water\n\nhow much water do you add to get to the weak, .6 mol/liter concentration\n\n12 mol/ (1+x) liters = .6 mol/liter\n\n12/(1+x)=.6\n12=.6(1+x)=.6+.6x\n11.4=.6x\n19 = x\n\nso if you had a liter of the solution at 12 M, you would have to add 19 liters of water and then you would have 20 liters of solution at the diluted .6 M\n\nbut you don't need that much of the final solution, so you use the same ratio of water addition, but only start with enough concentrated to end with 300 ml\n\nthe ratio is 19 to 1\n\nso 1/20 * 300 = 15   you need 15 ml of the conc. solution\n   19/20 * 300 = 285  you add 285 ml of water \n\nthe total is 300 ml of the .6 M solution\n\nnow look back at the equations and you will see where they came from and how to apply them properly
1098900
The specific heat at constant volume for solid KCl as a function of temperature between 298 and 900 K is given by  (see 1st source):\n\nCp° (J/mol/K) = A + B*t + C*t^2 + D*t^3 + E/t^2\n\nwhere \n\nA = 35.41597 \nB = 70.03472 \nC = -91.38233 \nD = 52.52426 \nE = 0.153460 \nF = -449.1357 \nG = 108.9982 \nH = -436.6841\nand\nt = Temperature in K/1000\n\ndelta-H of formation of KCl at 298.15 = -436.68 kJ/mol (1st source)\n\ndelta-H of formation of KCl at 298.15 = -436.47 kJ/mol (2nd source\n\ndelta-H of formation of K+ (aq) at 298.15 = -252.17 kJ/mol (2nd source)\n\ndelta-H of formation of Cl- (aq) at 298.15 = -167.08 kJ/mol (2nd source)\n\ndelta-H of solution (using data from second source for KCl(s)):\n\n-252.17 +(-167.08) -(--436.47) = +17.22 kJ/mol\n\ni.e., this is the enthalpy change for the reaction:\n\nKCl (s) -> K+ (aq) + Cl- (aq)\n\nCan't help you with the value for Cv, but if you happen to know the molar volume, molar expansivity, and isothermal compressibility of solid KCl, at the temperature of interest, you can calculate Cv from the thermodynamic relationship:\n\nCp - Cv = T*V*a^2/K\n\nwhere T is the absolute temperature, V is the molar volume, a is the isobaric expansivity, and K is the isothermal compressibility.
342141
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
446321
Hmmm, a multi-step problem!  Love those!\n\nFirst, you will need to know how many moles of oxygen are generated.  For that, you will need to use the Ideal Gas Law:\n\nPV = nRT\n\nYou need to solve for n = PV/RT, and since you know the volume, pressure, and temperature, and the value of R is a constant, this is an easy calculation.\n\nThen, from the equation:  \n\nKClO3 ===> KCl + 1.5 O2\n\nyou can work backwards to determine the number of moles of KClO3 you started with (one mole of KClO3 generates one and a half moles of oxygen, thus the number of moles of KClO3 you started with will be 2/3 of the number of moles of oxygen generated).  Knowing how many moles of KClO3 you had, you can then determine the starting mass of the KClO3, then find out what percentage that was of your original starting mixture.\n\nGood Luck!
529314
The molarity of a solution is the number of moles of a substance per liter of solution.\n\nMolarity = moles/liter\n\nSo, if you know what your solute is, you need to calculate the number of moles you have by:'\n\nmoles = mass of solute/molecular weight of solute\n\nnumber of moles is then divided by the number of liters you have, and voila!  Molarity.\n\nExample:  40 grams of sodium chloride in 500 mL of water:\n\n40 gm NaCl/58.443 grams per mole = 0.6844 moles\n\n0.6844 moles/500 mL * 1000 mL/liter = 1.369 moles/liter  or the solution is 1.369 M sodium chloride in water.
554942
Q =  m C (T-t)\nQ = (2kg)(2100j/kg)(0°c-(-20°C))\n= 84000 j\nthen\nQ = mL\n= (2)(335,000)\n= 670000 j\n\nthen\nQ = m c (T-t)\n= (2)(4200)(100-0)°C\n= 840000\n\nthen\n\nQ = mL\n= (2)(2,260,000)\n= 4520000\n\nthen\nTotAl = 8400+670000+840000+4520000\n\n\n= 6,038,400 J
272997
ok i will TRY to help you. im not completely sure but i think i know what to do. basically, when you add K2Cr2O7, it splits up into 2 parts, the Ks and the Cr2O7s. the Cr2O7s react with the potassium dichromate and make products. so if it was titrated correctly (assume it is), then \n8.76mL * 1L/1000mL * 0.04988mol/L    mols were added\n= 0.0004369488 mol of Cr2O7 added. and this amount made all the potassium dichromate get used up (apparently).\nwhich means that 0.0004369488/2 mols of potassium were in the blood. (divided by 2 because for each mol of potassium, 2 mols of Cr2O7 were used). so 0.0002184744 mols of potassium were in the blood. and if you use the molar mass to convert that to a mass. and then divide it by 10.0002g. you should get the mass percent of blood alcohol. so 0.0002184744mol x [(2x12.01) + (4x1.01) + (1x16) + (1x1.01)]\n= 0.0002184744mol x 45.07 g/mol\n= 0.009846641208 g  and divide it by 10g to get the mass percent which is 0.098466%.\nhope this helps. let me know if u need more help. laterz
179716
Here are the equations you need to know:\n\nMolarity (M) = moles/Liters\n\nc=(t 1/2)(2k); \nor concentration  = half life (t 1/2) * 2 * rate constant (k)\n\nUsing this equation, you can solve for the rate constant, k.\n\nThen you can plug that value into the equation for a first order rate law:   \n\ncf=c1-kt; \nor final concentration = initial concentration - rate constant * time\n\nGood luck with the calculations!
1051997
This is the most handy of chemistry equations, if you need to take more chem, commit it to memory:\n\nm1v1=m2v2, molarity*volume=diluted molarity*diluted volume\n\nSo, 9.9M*.102=x*.525.  Then solve for x by dividing the left side by .525.  You will get the concentration in mol/L of the final solution.\nSorry, I read your question wrong, I read it as diluted to, not diluted with, so you just need to add the volumes.\nSo it's 9.9*.102=x*(.102+.525)\n9.9*.102/.627=concentration of 1.6\nSorry, I wasn't more carefull.\nThat really is the way to solve these equations though.