What is the conductivity of a 50% HF solution in water?
469936
quite high but the HF will dissolve the electrode lol
810698
1. VERY PURE water is a poor conductor of electricity. However, very small \namounts of impurities, especially salts, greatly increase the electrical\nconductivity of water.\n2. The main carriers of electrical current are dissolved ions. If a \nbattery is connected to a container of water so that\nthe (+) and (-) poles of the battery are immersed in the water, the\npositive ions will migrate toward the (-) pole and the negative ions will\nmigrate toward the (+) pole (electrical opposites attract). This closes the\n"switch" and electric current flows through the solution.\n3. Different ions can help carry the electric current, some more / some \nless, depending upon their electrical charge and their size. More highly charged, \nsmaller ions are more efficient carriers of the electrical current.\n4. The higher the concentration of ions, the greater is the conductance of \nthe solution.\n5. There is a complication however, the water and / or certain of the ions \nmay undergo a chemical reaction to one or the other, or both of the electrodes.\nYou may not want to introduce this in the beginning but it does happen.\n6. None of the ion electrical current carriers of salts dissolved in water \nare as mobile as "free" electrons in a metal are. And the differences are very\nlarge. For example, if the conductivity of  VERY PURE WATER is 0.000004\nconductivity units, the conductivity of a saturated solution of common\ntable salt NaCl is 23 conductivity units, and the conductivity of a typical\nmetal such as copper wire is 60, 000, 000 conductivity units. So while the\nions in a dissolved salt behave something like "free" electrons, they are\nvery slow and sluggish moving. \nCompletely demineralized water actually does not conduct electricity, and is\nwidely used as a coolant in high powered radio amplifiers.  However, when\nexposed to even small amounts of impurities, (even normal air), It becomes a\nfair conductor.  It is not the water conducting this electricity so much as\nthe dissolved minerals in it.  Some of these minerals are decent insulators\nthemselves, but when dissolved in water still allow conduction because the\nmolecules are now free to move around.\nIn ionic liquids these electrons remain \nstuck to big, round atoms,\nand these atoms do the swimming around, relatively slowly, dragged by \ntheir electric charge.\nSort of a ball and chain thing.\nElectrons in metal are "free"; they slip rather freely across all the \natoms, and do not need to drag one around with them.\nSo, carrier mobilities in ionic liquids are some orders of magnitude lower \nthan carrier mobilities in metals.\nSo their resistance is higher.\nIf you had a snail-trail of salt-water or molten salt with the same length \nand cross-section as a metal wire,\nit would use up half the voltage and get hot trying to pass one tenth the \ncurrent the wire can handle.\nIn a metal, each metal atom contributes about one mobile electron that \nhelps the current flow.\nIn molten salts like NaCl, every atom has a + or - charge and contributes \nto the current flow.\nSo there are just as many carriers per unit volume as in a metal.\nBut in water-solutions of salts, the salt is a minority.\nUsually less than 10%  of the solution is ions, so that makes the \nconductivity even poorer.\nThat is part of why we will never make our power lines out of seawater in \npipes.
561926
XeF  + e-  ============> Xe  + F-      E°/V = 3.4 with reference to the standard hydrogen electrode
721562
Beryllium sulfate, in the anhydrous form, absorbs water to become the tetrahydrate immediately upon contact.\n\nThe solubility of the tetrahydrate form in cold water is 42.5 grams/100 mL of water at 25°C.  The solubility in boiling water (100°C) is 100 grams/100 mL water.
786992
b) small fraction of the maximum number of possible hydronium ions, H3O+.\n\nHNO2 + H2O <===> H3O+ + NO2-
801826
E=hf  where h = 6.6210^-34 Js\n\ne = 6.6210^-34*4.31*10^14 = 2.71 10^-19 Joule=1.7 eV
432258
Since it is at thermal equilibrium, the energy (q) of the water must equal the energy (q) of the metal.\n\nTherefore: mcAT = mcAT\n\n(90)(4.18)(0.9) = (11)(87.2)(c)\n\nSolving for c, the specific heat of the metal would be 0.353 J/g*oC
351009
Water with a high salt concentration, and plasma (electrically charged gas)
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Hydronium ions, in an aqueous solution.  H30+
136403
You can calculate the freezing point depression of the solution is you know three things, firstly, the freezing point of the pure solvent, and secondly, the concentration of the solute, and thirdly, the freezing point depression constant of the solvent.\nIn this problem, you know both (One is given, two of them are assumed knowledge).\n\nThe freezing point of pure water is 0 degrees Celcius.\nThe concentration of the solute is 52 percent.  You need to convert this to a molaltiy concentration to use it (or fiind a different freezing point depression constant).\nLook up for your self the k_f value.\n\nthe freezing point depression = k_f * m,\nwhere k_f is the freezing point depression and m is the concentraion of the solute in terms of molality.\n\nmolality = moles of solute / kg is solvent
910705
Hello, here is what you need:\n\nEnergy = Planck Constant * Frequency\n\nE = h * f\n\nh = 66.26 x 10^-34 Joule second
1167951
HgCl2 - 0.48 mol / 100 mol water at 25 degrees C\nHg2Cl2 - very insoluble in water
1108709
Safe to drink... or safe to be around in the presence of an electrical current?\n\nWell zero conductivity would be ideal, because it would be evidence of zero ion presence in the water.\n\nHowever the water could be very conductive and still be "safe" to consume. Salt (Na-Cl) for instance mixed into the water would make turn the water into an ionic solution. It is therefore now has conductivity. But as we all know... we can consume salt water "safely" if it is consumed at appropriate amounts.
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
573262
Pure water is *not* a good conductor of electricity.  Dissolved ions are necessary to conduct electricity and at room temperature, there is only 10^-7 mol/L of H+ and OH- in pure water.\n\nDissolving salts (including strong acids) in water greatly increases the conductivity of the solution.\n\nTap water is a modest conductor of electricity because of dissolved ionic minerals and impurities.\n\nHeating water increases its dissociation constant making more H+ and OH- available in solution.
573821
I think this question is incomplete because the Ka of HF hasn't been provided.  The amount of H+ that is formed from HF can only be calculated from its Ka value.
768198
molarity =moles/V where volume is expressed in L.\nmoles=mass/MW so\n\nmolarity=mass/(MW*V) =50/(106*0.5)=  0.94 M\n\nFor solutions, % (w/v) is grams of solute per 100 mL of solution.\nThus you have (50/500)*100% = 10%\n\n% is parts per hundred.\nso it is 10^4 times bigger than ppm\n so it is 10*10^4 ppm= 10^5 ppm.\n1 ppm corresponds to 1 mg of solute per 1L of solution.