How do you calculate the new molarity that results when 250 mL of H2O is added to 125 mL of a 0.251 M HCl sol?
341279
125mL of 0.251 M HCl gives 0.125 * 0.251 mols of HCl\n\nThen 250 + 125 mL is 375 mL = 0.375 L\n\nSo 0.125 * 0.251 mols in 0.375 L gives 0.125 * 0.251/0.375 M\n\nThat is 0.084 M
280181
ok so.  first   balance the equation. shouldnt be too hard. 2Cr  3Cl2   and     2CrCl3 \nchromium u dont have to worry about cuz itz excess. you have 125g of Cl2 though. so you have to find   out how many moles that is. by dividing by the molar mass. so once you find out the   number of moles reacting  and then  find out the according number of moles of CrCl3    that are formed. in this case  its the same number    so    then   once you know the number of moles. you    find the   mass by  multiplying by the molar mass. and there u go
165367
molality = moles of solute / kg of solvent.\n\n200 grams of water  = .2 kg of solvent.\n\n.3 molality = x moles / .2 kg\nsolve for the number of moles (x).\n\nThen multiply the number of moles by the Molar mess of Silver Nitrate and that is your answer.
681504
For this you need to the density or molarity of the solution.\n\nUsing molarity:\nMolarity=(Number of moles of solute)/(litre of solution)\n\nAlso 1 litre=1000 cubic centimeter.\n\nUsing density, d=m/V you get the mass of the solute.\nFrom there you can go to moles by using \nmoles=mass/(atomic or molecular mass)
1115475
To make 100mL: \n\n0.6% from a 10% solution is 6ml of the 10% diluted in 94ml of H2O.  \n\n0.6% from a 25% solution is 2.4ml of the 25% diluted in 97.6ml of H2O.\n\nThe math?\n\n0.6%*100mL / 10% = 6ml\n\n0.6%*100mL / 25% = 2.4ml
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.
773826
the answer that i got is 3.10 M. \n\n3.25 g NaOH/40.00 g NaOH = .0813 mol NaOH\n\n26.2 mL / 1000mL = .0262 L\n\nM= .0813 mol / .0262 L = 3.10 M
952447
I think you probably mean that the half liter is mixed with 250 ml. Otherwise the volume afterwards couldn't be 750 ml. Okay, the amount of HCl in the half liter is\nn=V*c=0.5l*2.5 mole/l=1.25 mole (remember that M simply means mole per liter).\nThe amount of HCl in the 250 ml is\nn=V*c=0.25l*3.75 mole/l=0.9375 mole\nThe total amount of HCl is then 1.25 mole+0.9375 mole=2.1875 mole meaning that the concentration in the final solution is\nc=n/V=2.1875 mole/0.750 l=2.92 mole/L=2.92 M.\n\nSince HCl is a strong acid it is completely separated in H+ and Cl- in the solution meaning that\npH=-log(concentration of H+)=-log(concentration of HCl)=-log(2.92)=-0.5.\n(Strictly speaking this formula for pH is only valid in dillute solutions like 0.1 M and below, but it is the best estimate we can give without making this problem VERY compliated).
685990
HCrO4- ---------------> H+   +  (CrO4)2-\nThis is an equilibrium problem, assume that x moles of HCrO4- dissociate which means that x moles of H+ and   (CrO4)2- are produced.\n\nAt equilibrium, \n[HCrO4-] = 0.250M-x\n[H+] = x\n[(CrO4)2-] = x\n\nKa = [H+] * [(CrO4)2-]/ [HCrO4-]\n3.0*10^-7 = x*x/(0.250-x)\n3.0*10^-7 = x^2/(0.250-x)\nFor these kind of problems you have to assume that\n0.250-x = 0.250\nTherefore, \n3.0*10^-7 = x^2/0.250\nx^2 = 7.5*10^(-8)\nx = 2.74*10^(-4)\n\n[H+] = [(CrO4)2-] = x = 2.74*10^(-4)\n\nPercent ionization = x/[HCrO4-] * 100\nPercent ionization = 2.74*10^(-4)/0.250 * 100 = 0.1095%\nor approximately 0.11%
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
786849
Molarity (M) = no. of moles(n)/ Volume of the solution(V) in litres\ni.e., M=n / V\n\nnow u need 3 moles\nHence n=3\nfrom 2 M solution\nTherefore M =2\n\nNow\nV = n / M = 3 / 2 = 1.5 L
832170
Molarity = moles of solute/liters of solution\n15.8 g of HF = 1.01 + 19 = 20.01 g/mol\n15.8 g HF / 20.01 g/mol HF = 0.790 mols of HF\n0.790 moles of HF / 0.200 L of solution = \n3.95M solution
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!
609634
This question cant be answered without knowing the molarity of the HCl. I seriously doubt you were using 12M as another poster suggested. \n\nAlso. if all the Mg reacted the moles of Mg reacted would simply be 0.15/24.3 = 0.006. (Did you really use 0.15g? Thats a small amount) \n\nIf you didnt actually preform the rxn and/or dont know if all the Mg reacted you need to use the moles of Mg calculated above, and compare it to the moles of HCl used. M = moles / liters. So plug in the molarity of HCl used, and 0.050 L (converted 50mL) and solve for moles of HCl. Next use the mole ratio to determine the limiting reagent. If you dont know how to do that, your screwed.
697635
1) Calculate the number of moles of NaOH in 300 mL\n3.81 g /( 22.99+16+1.008) = 0.0955 mol \n\n2) Calculate the number of moles of NaOH in the 25 mL transfered\n0.0955 mol * 25/300 = 7.96 x 10^-3 mol\n\n3) Calculate the molarity \n7.96 x 10^-3 mol / 100 mL = 0.0796 mol/liter
365444
0.500L x 0.250moles/L x 58.44g/mol = 7.31g\n(a)