If I prepare a water solution of Baking Soda and a dilute solution of Vinegar, will both solutions react slower and over a longer period of time than if I just put baking soda directly into vinegar ?\n\nHow may I calculate the reaction rate ?\n\nThis is basically a concentrations VS reaction rate problem.
133697
yes, reaction rates are based on several things, concentration and temperature are the most important, because of the water added to the reaction, it will be less "likely" that a vinegar molecule will collide with a baking soda molecule, because of this the reaction will go slower. the way to calculate this is K = [Products]/[reactants].  Where [] is the concentration of each solute present (in moles per liter), and K is the constant value for this perticular reaction, which can be found in a chemistry text book.
479190
For reaction aA <--> bB + cC, the Eq.constant K is defined as\n\nK = (B^b * C^c) / A^a\n\nSince in your equation a=b=c=1, and concentration for B=C, the equation simplifies to\n\nK = B^2 / A\n\nwith B=0.046 and A=0.454    (0.454=0.500-0.046)\n\nso K = 0.046^2 / 0.454 = 0.0047
429913
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
365270
The vinegar solution is hypotonic to the egg, the corn syrup is hypertonic to the egg and the water is isotonic. This is why the vinegar reacts with the egg (removes ions from it's shell) but the corn syrup and water will not.
448049
Yes, it is true that the density of a solution is different of solvant but tjis difference is small ; So , i propose to dissove 40g of amyl acetate in a liter of water if that acetate is a solid. If that acetate is aliquid the total volume will be 1.04 l and you must use 47 cc tp water
251829
Assuming they are mixed:\n\nFirst, the oil should separate from the vinegar water and sugar and float on top.  You could siphon or pour that out.\n\nNext, you could heat up the water, vinegar, and sugar solution slightly.  This should cause the vinegar to vaporize first (it could be the water, you would have to look that up first) and that should separate the vinegar, and the sugar and water would be left.\n\nThen heat that up even more, boiling the water, which will leave the sugar.\n\nIf you still need everything intact, you would have to use a condensor system to collect the vapors and cool it to return to a liquid form.
136809
You sure did pick a really long question for explaining.  And I'll have to argue that all of the pH's do not have correct sig figs but the numbers are right.\n\n(Sorry about the numbers in the tables but this place doesn't seem to allow better tables)\n\nA) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH\n\n*This is a reaction so we have to use moles*\n\nHAc : 25.0mL * (0.100mol/L) * (1L/1000mL)= 0.00250 moles\n\nNaOH : 10.0mL * (0.125mol/L) * (1L/1000mL) = 0.00125 moles\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00125      0.00250        0\nChange:   -0.00125     -0.00125      +0.00125 \nEquilib:      0         0.00125       0.00125\n\nHAc and Ac-(NaAc breaks apart) for an equilibrium\n*Use Molarity when no reaction is taking place (best shown by when no volume is changing)*\n\nBoth HAc and Ac- have same molarity:\n\n(0.00125 moles / 35mL ) * (1000mL/1L) = 0.0357M HAc and Ac-\n\nUsing the equilibrium equation:\n\nEquation:  HAc   <--> Ac-     +  H+         Ka = 1.8 * 10^-5\nInitial:   0.0357M     0.0357M  \nChange:    -x          +x         x\nEquilib:   0.0357-x    0.0357+x    x\n\nKa=1.8*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]\nKa=1.8*10^-5 = [0.00357][x]/[0.0357] usually good approx.\nKa = 1.8 * 10-5 = x = [H+]\n\npH = -log [H+] = -log [1.8*10^-5] = 4.745 (3 sig figs as only after the decimal counts for logs)\n\nB) 25.0mL HAc and 20.0mL NaOH\n\n*See A for HAc moles and double the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00250      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:      0            0          0.00250\n\nCalculate molarity of Ac-:\n\n(0.0250moles /45mL) * (1000mL/1L) = 0.0556M Ac-\n\nUsing the other equilibrium equation (Kb):\n\nEquation:  Ac- + H20   <--> HAc     +  OH-         Initial:   0.0556M   -        0\nChange:    -x        -        +x        + x\nEquilib:   0.0556-x  -       +x         + x\n\nThis gives us the Kb equation\nKb = ([HAc][OH-])/[Ac-] = with approx ([x][x])/[0.0556]\nso Kb = x^2 /0.0556\n\n**but we need to calculate Kb\n**Kw = 1*10^-14 at STP I'll assume this for this problem\n**Kw = KaKb so Kb = Kw/Ka\n**Kb = (1*10^-14)/(1.8*10^-5)= 5.56*10-10 \n\nso x = 5.56 * 10^-6 = [OH-]\n\npOH = -log [OH-] = 5.253\nPH = 14 - pOH (at STP)= 8.747\n\nC)25.0mL HAc and 30.0mL NaOH\n\n*See A for HAc moles and triple the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00375      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:   0.00125            0       0.00250\n\nAfter the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.\n\nCalculate NaOH molarity\n(0.00125moles/55.0mL)*(1000mL/1L)= 0.0227M NaOH\n\nStrong Bases dissociate 100% so [OH-] = 0.0227M\n\npOH= -log[0.0227] = 1.644\npH = 14-pOH = 12.356\n\nThis should show how they calculted the half-neutralization point (A), the neutralization point (B) and after the neutralization point (C).\n\nIf you have questions on this, feel free to send me a message.  I taught this a couple weeks ago so I'm still pretty fresh on the subject.
509259
I did a demonstration on the affects of acid rain on plants once.  I used 5 identical, healthy house plants and watered them everyday with varying ratios of a water:vinegar mixture.  Pure water, 5 parts water:1 part vinegar, 5:2, 5:3, 5:4 and so on.  I recorded my observations each day for 10 days.  The impact that the acidity had on the plants with high vinegar ratios was pretty neat to watch and document.  You of course would have to explain how the vinegar acted like acid rain would but there are plenty of resources for this on the net.
479683
A thorough perusal of your textbook should help with your problem.  Oh, wait--sorry, I forgot--you aren't interested in LEARNING how to calculate the pH of your sodium acetate solution, just in getting the quick fix answer from someone who has.
181824
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
80985
If you drop a tooth into concentrated acid (e.g. sulfuric acid) nothing much happens. You might expect a bit of a fizz, but this is not the case. The tooth has to be in the acid for quite some time before any difference is noticeable.\nThe acetic acid in vinegar is very dilute, so even if you rinsed your mouth with vinegar it would not harm your teeth.\nThe amount of vinegar in food is harmless, and in any case is less acidic than the normal stomach acid (hydrochloric acid) anyway.\nVinegar from our food gets nowhere near our bones.
363632
You don't give the rate law, so it's impossible to tell. Suppose [H2O2] = concentration of H2O2 and [I-] = concentration of iodide. If rate = k[H2O2][I-], then the mechanism is one step. This means that both an H2O2 and an I- have to come together at the same time for the reaction to occur. If rate = k[H2O2], then it's a two-step process. This is because something has to happen to H2O2 in a slow, rate-limiting step, followed by the reaction of whatever with I-.
419805
Neutralization Reaction:\nHC2H3O2(aq) + KOH(aq) -->H2O(l) + KC2H3O2(aq)\nThe equation is already balanced.\n\na. False KC2H3O2(aq) and H2O are products.\nb. True\nc. False, some ions will cancel out but some will remain.\n\nThe ionic equation is as follows:\n(H^+)(aq) + (C2H3O2^-)(aq) + (K^+)(aq) + (OH^-)(aq) -->H2O(l) + (K^+)(aq) + (C2H3O2^-)(aq)\nRepeated ions are crossed off from both sides and the net ionic equation is:\n(H^+)(aq) + (OH^-)(aq) -->H2O(l)\n\nd. True\ne. False\n\nNote: the + and - signs were written with ^ to indicate that they were raised up (like a power).  Sorry if this notation is confusing.  Good luck and I hope that this helps!
258064
Well, the equation is..\nCH3COOH + NaHCO3 ----> CH3COONa (Sodium Acetate) + H2CO3 (Carbonic Acid) \n(or)\nCH3COOH + NaHCO3 ----> CH3COONa (Sodium Acetate) + H2O (Water) + CO2 (Carbon-dioxide)\n\nSo you might have noticed the equation is already balanced. It is a double-displacement reaction. So the sides are already balanced.\nHope this helps.\n\nOne small note...I don't think there is a compound such as NaCO3. The valency of CO3(^-2) ion is 2, so it has to combine with two Sodiums (two positive charges to neutralise its two negative charges) unless the compound one of the above users mentioned is NaCO3-.\n\nrule wid style n luv@vinny
372854
Its an acid base reaction. The baking soda (sodium bicarbonate) is a base and neutralizes the acid in the vinegar (acetic acid) and releases carbon dioxide in the process.\nHere is the reaction :\nNaHCO3 + CH3COOH ———> NaCH3COO- + H2O + CO2
105653
if we are talking just Vinegar (acetic acid) and sulfur there is a possibility it would react to form hydrogen sulfide (H2S). H2S is toxic so you need to do it in small amounts and a WELL VENTILATED AREA!!!
421489
Are there any tests you can run? Usually if you mix a substance with something else, the product has different chemical and physical properties than the initial substance.\nFor example, let's say that you would mix isoamyl acetate with "X" the product will have certain carachteristics gained only after it was mixed with "X".... I don't know how much this will help...\n\nLook it up on the internet, I am sure you'll find something.\nGood luck.