Can anyone explain the Na-K ion exchange theory(sodium potassium ion exchange theory)?
456987
The concentration differences between K+ and Na+ across cell membranes create an electrochemical gradient known as the membrane potential. ATP provides the energy to pump Na+ ions out of the cell in exchange for K+ ions. This is done to correct cellular concentration.
176528
http://www.chem1.com/CQ/ionbunk.html\n\nskim this article.
804535
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!
919726
Sodium and Potassium are correct.  They are extremely reactive with water, forming Hydroxides and releasing Hydrogen gas, which is usually ignited from the exothermic heat of the reaction.\n\n2 Na + 2 H2O -> 2 NaOH + H2\n\n2 K + 2H2O -> 2 KOH + H2
808333
You have a common ion effect problem\n\nPbCl2 <=> Pb+2 +2Cl-\n\nKsp=[Pb+2][Cl-]^2\nYou know that [Pb+2]=5.5*10^-3\nAlso from the reaction you know that the amount of Cl- coming from the PbCl2 is double the amount of Pb+2.\nSo if C is the concentration of Cl- coming from HCl, the amount of Cl- at equilibrium is [Cl-]=C+2*[Pb+2] \n\nSo Ksp= [Pb+2]* (C+2*[Pb+2])^2=>\nif we assume that C>>2[Pb+2] then C+2*[Pb+2] =C and \nKsp=[Pb+2]C^2 =>1.7*10^-5= 5.5*10^-3 C^2 =>\nC=55.6 * 10^-3 but 2*[Pb+2]=11*10^-3, so we can't do the approximation...\n\nLet's name [Pb+2]=p for simplicity\n\nKsp=p(C+2p)^2 =>\nKsp/p= C^2+4pC+4p^2 =>\nC^2+4pC+4p^2-Ksp/p=0 =>\nC^2+2.2*10^-3 C -2.97*10^-3 =0\n\nC=0.5*(-2.2*10^-3 + SQRT( (2.2*10^-3)^2 +4*2.97*10^-3) ) =>\nC= 5.34*10^-2\n\nIf you added Va L of HCl, then\nC=mole HCl/Vtotal = Ca*Va/(Va+0.12) =>\nCVa+0.12C=CaVa =>\nVa=0.12C/(Ca-C)= 0.12*0.0534/(0.12-0.0534) = 0.0962L or 96.2 ml
976855
Ions are basically charged atoms. The formula of an ion can be found out if you have the valency of the atom. You can figure out the valency of any atom by finding the group at which the atom is located in the periodic table. For example, Sodium (Na) is located in group 1A of the periodic table. Hence the valency is +1. The formula of the sodium ion would be Na+1.\n\nHope that helps!
644652
I answered this already.\n\nhttp://answers.yahoo.com/question/index;_ylt=AmW0Em3eAdmFrnQMc4_Xilzsy6IX?qid=1006051323791\n\nWorks the same as if you were changing permeability. If you have more Na inside the cell, there is less drive for Na outside the cell to come in. Because Na only has a minor impact on the resting membrane potential, it won't change much, but it will hyperpolarize a bit (away from the Ena and towards Ek).
163022
Ion exchange is a viable alternative but it ultimately works on the same principles as Electrochemistry
311730
I think you should keep it simple yet smart. First get a piece of pure elemental Sodium and then let it drop on warm water. Afterwards explain how the chemical reaction released Sodium Oxide and Hydrogen Gas.
729404
They yield K3PO4 and H2CO3.... Just switch the positive ion with the negative ion and make sure the charges cancel out to zero on each new salt.
382937
While a cell is in it's resting membrane potential, the amount of K+ inside the cell is greater than extracellular potassium; at the same time, Na+ is in higher concentration outside of the cell than inside. While in this "resting" state, the cell is more negative inside the cell due to the mechanism of the Na+/K+ pump. 3 Na+ are pumped out of the cell while 2 K+ are pumped into the cell. Since both Na+ and K+ have plus 1 charges, the cell inside will be more negative than outside because of the difference in charge. \n\nWhen the cell is "excited", something called sodium gated channels open and Na+ shoots into the cell, making it more positive. This increase of positive charge shoots past the threshold potential (the amount of energy required for the action potential to be initiated), and the action potential is passed. Once it reaches a peak, potassium channels open, causing the cell to become more negative once again inside the cell. As soon as the potential goes back to it's resting potential, the cell is ready for another response.
954955
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
836198
OK....You need to understand how this is all working, and why.  So, the formula equation for this reaction would be something like:\n\nKOH(aq) + HNO3(aq) --> H2O(l) + KNO3(aq)\n\nSince KOH, HNO3 and KNO3 are all strong electrolytes, they are totally ionized when they're dissolved in water.  So, written in complete ionic form, that equation would be:\n\nK+(aq) + OH-(aq) + H+(aq) + NO3-(aq) --> H2O(l) + K+(aq) + NO3-(aq)\n\nTo get to the net ionic equation, you just cross out the things that are the same on both sides of the arrow.  K+(aq) and NO3-(aq) are on both sides, so the net ionic equation is:\n\nH+(aq) + OH-(aq) --> H2O(l)
740901
You need to split the ionic compounds apart to find the individual ions you would find in these structures.\n\nMgO would produce Mg^2+ & O^2- ions.\n\nK2S would produce K^+ & S^2- ions.\n\nYou can find the charge values directly from the periodic table trends for the charges atoms assume when they are ionized.
351293
it is based on arrhenius equation\nK=A e ^ (-Ea/RT)\nA- frequency factor\nEa- activation energy (proportional to Delta H)\nR -Gas constant\nT -Temperature\nSince you dont know A- fequncy fator, yu need to do a lengthy calculation\n\nThe modified arrhenium equation will be ( by considering two situations T1, T2 there will be two rates k1, k2 but A-frequency factor is smae for one reaction so it's a constant..)\n\nEa= (R X T1 X T2 X Ln (k1/k2) ) / (T1-T2)\n\nEa= delta H = 17,390 J\nR = 8.314 J/K\nK1 = 1.6 X 10E-4\nK2= ?\nT1 = 298\nT2= 273+97= 370K\n  After substituing all these values, you will get K2= 6.27 E-4\nwhich is in agreement with theory (At high Temperatures, rate should be high..!)\n\nIs this info will help you?\nGood luck!
551370
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
943195
It has been many years since I even cracked the book on crystal field theory, but I refreshed myself a bit here:\n\nhttp://www.unine.ch/chim/chw/Chapter%204.html\n\nBasically, the difference in splitting (delta value) decreases as you go down a group and increases by about 50% as you add positive charge on a metal cation. Other factors are covered on that page as well. Thanks for the reminder that I took a graduate inorganic chemistry class once back in the early 90's! I assure you that, at one time, I actually understood it, but I haven't used it in so long...well, you get the picture!\n\nP.S. (I had forgotten that it replaced VB theory, which I was supposed to teach my students but ran out of time!)