Comparing 45cm3 of 1.2 M HCL  And  60cm3 of 0.9 M CH3COOH. Which one will have a higher pH value?
210887
45 cm3 = 45 mL\n\n45 mL * 1.2 M HCl = 54 mmol HCl\n\n60 mL * .9M HAc = 54 mmol HAc\n\nSince the material amount is the same, HAc (acetic acid, CH3COOH) will have a higher pH because HCl is a strong acid and it will have a low pH. CH3COOH is a weaker acid, so its pH would be higher. HCl is a strong acid because the H and the Cl are strongly solvated by water molecules because of the immense polarity of the H-Cl bond. CH3COOH is weak because, while the anion is resonance stabilized, you still have a molecule with a hydrophobic end and an electron donating group. (the CH3)
536609
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
443749
The greater the Kb value is, the stronger the base.  Just think about what Kb means.  On the left side of the equilibrium is the base, on the right side is the conjugate acid, meaning the base has reacted.  The stronger the base, the more it will react, the more the reaction will proceed to the right, the greater Kb will be.\n\nSo.....\n\nC > E > B > A > D
315350
A 30% Acetic Acid solution has a density of 1.05 g/mL.  Water has a density of 1.0 g/mL.  So it is reasonable to assume that Acetic acid ("vinegar") is more dense than water, but the two are very, very close indeed.\n\nWhen you dissolve Acetic acid in water, the two mix to form a solution.  It is not easy to distill away the acetic acid from the water.\n\nJust because something is more/less dense than water does not necesarily mean it has a lower/higher boiling point.
422833
M=molesof solute/Liter of solution\n\nM= molarity
439538
Boyle's Law: P1V1 = P2V2\n\n40. (65.0mmHg)(600mL) = (385mmHg)(x)\n      39000 = 385x\n      101.3 = x\n\n41. (1)(2.85) = (x)(1.70)\n      2.85 = 1.70x\n      1.68 = x\n\n\n\n...basic algebra
445741
If tha atomic mass of M is A, then the molecular weight is MW=2*A+60\n\nThe concentration of the M2CO3 solution is C=mole/V and since mole=mass/MW, C=mass/(MW*V)\n\nThen you do a titration for which we assume complete reaction and not reaching only the first end-point, so the reaction is\n\nM2CO3 + 2HCl -> 2MCl + CO2 + H2O\n\nTherefore 2*mole M2CO3=mole HCl =>\n2*C1*V1= C2*V2=>\n2*V1*mass/(MW*V) = C2*V2 =>\nMW= 2*V1*mass/(C2*V2*V)=\n=2*0.025*7.46/(0.1*0.027*1)\n=138\n\nand MW=2*A+60=138=> 2A=78 => A=39\n\nSo M, is potassium (K)
491261
ph=-log[h+]\n    hence h+ conc.= 10^-2.5=0.0031 moles/l\n    0.5m= 0.5 moles/l\n% = 0.0031/0.5*100\n     =0.632%
284154
The first answer is poh=3.167 and the ph=10.832\nThe second answer is poh=2.119 and the poh=11.881\n\nNot sure onthe concentration of hydrogen though.
490261
Ba(OH)2 + 2CH3COOH ----> (CH3COO)2 Ba\n\nmole CH3COOH: 12g/ 60g/mol = 0.2 mol\nmole Ba(OH)2: 0.2 mol/ 2 = 0.1 mol\n - since 2 mol acetic acid is required for every mole Ba(OH)2\n\nIf concentration of Ba(OH)2 is 0.1M, volume required to react is 0.1 mol/ 0.1 M = 1L or 1000 cm3
476390
The pH of the salt depends on the strengths of the original acids and bases:\n\nAcidBaseSalt pH\nstrongstrongpH = 7\nweakstrongpH > 7\nstrongweakpH < 7\nweakweakdepends on which is stronger\n\nThat should be what you want.
413987
4PH3 (g)  ---> P4(g) + 6H2(g)\n\n4 moles of PH3 are consumed then 1 mole of P4 is produced and 6 moles of H2 are produced.\n\nIf you are consuming 0.0048 mol/s of PH3, then by the relation above you will be producing:\n\n0.0048/4 = 0.0012 mol/s P4\n0.0048*6/4 = 0.072 mol/s H2.
274662
This is a typical sort of problem found in real life laboratories- You have differences in concentrations because, in the real world, there are errors introduced at every step of the quantitation process..tiny differences in how much stays in the tube, in the pippette tip, the accuracy of the pippette itself, etc.  However, figuring out which is more accurate is fairly easy.  First, any value that actually falls *on* your standard curve is best-that way you don't have to extrapolate. Second, in general, the more you dilute, the less accurate the reading.  The first consideration is the more important.  In sample 1, your 1:50 dilution is too low to get an accurate reading, in unknown 2, your first dilution is too high to get an accurate reading.
466164
<--------- 7 ---------->   (pH)\nacid                base\n\nthe smaller pH the stronger acid\nthe bigger pH the stronger base\n\nso you should arrange them from the smallest pH to the biggest\nstomach content (1.6)\nsoda pop (3)\nmilk (6.5)\nfresh water (7, of course neutral)\nhuman blood (7.4)\nsea water (8.7)\ndrain cleaner (8.7)
419731
well I think no. because dilute HCl has a low concentration therefore it produces less hydrogen ions. so the less hydrogen ions it produces the less pH  it has
525814
11/27:Please post the answer so I can find the error in my work or determine if the answer key is wrong.  Lets see if we can get to the bottom of this one!\n\n\nFirst find the masses of each acid:\n\nHCl:  48.8 ml x 1.19 g/ml x 0.38 = 22.1 g\n\nHNO3:  17.5ml x 1.42 g/ml x 0.70 = 17.4 g\n\nSecond, convert grams to moles:\n\nMass HCl = 35.45 + 1.01 = 36.46 = 36.5 g/mol\nMass HNO3 = 16.0 x3 + 14.0 + 1.0 = 63.0 g/mol\n\nmoles HCl = 22.1 g/36.5 g/mol = 0.605 mol HCl\nmoles HNO3=17.4g/63.0 g/mol = 0.276 mol HNO3\n\nSince you're diluting to one liter, these are also the molarities.\n\n[HCl] = 0.605 mol/L ; [HNO3] = 0.276 mol/L\n\nEach of these acids are strong acids and can be assumed to dissociate completely.  If they were weak acids we would have to account for the acid dissociation equilibria here.  \n\nEach of these acids contribute only 1 equivalent of hydrogen ion.  Thus [H+] is the sum of the two concentrations.\n\n[H+] = [HCl] + [HNO3] = 0.605 + 0.276 = 0.881 mol/L = 0.88 M\n\nReport two sig figs since your mass % vals only have two sig figs.\n\nThe equilibrium equation for water, H+ and OH- is:\n\nH2O <--->  H+  + OH-   K = 1.00 x 10^-14\n\nH2O is a pure liquid and by definition has an activity of unity, so the equilibrium expression is:\n\n[H+][OH-] = 1.00 x 10^-14\n\nNow we can solve for [OH-] :\n\n0.881 X [OH-] = 1.00 x 10^-14\n\n[OH-] = 1.14 x 10^-14 mol/l = 1.1 x 10^-14 M\n\nAgain, two sig figs in final answer.\n\nFinally, pH = - log [H+]\n\nso pH = - log 0.881 = 0.055
295960
Look at their conjugate bases and whichever is more stable will have the weakest acid. PKA values are also easy to prove this becuase it is a realtion of the H+ content.