Relationship Between Kp and Kc?
375635
Kp=Kc(RT)n\na.Kp=Kc(RT)-1/2\nb.Kp=KcRT\nc.Kp=KcRT\nd.Kp=Kc(RT)2
497267
thatguybil has the explanation for competitive inhibition correct.  The Km will decrease as concentration of the inhibitor increases, but Vmax will generally not change.\n\nIn the case of uncompetitive inhibition, the apparent Vmax and Km will both decrease as the concentration of inhibitor increases.  Unlike noncompetitive inhibitors, which bind both the enzyme and the ezyme-substrate complex, uncompetitive inhibitors bind only to the enzyme-substrate complex.
335806
Because KOH is virtually completely soluble, there is no real Kb value for it.  A Kb value is calculated by multiplying the concentrations of the ions that KOH forms in solution, then dividing that number by the concentration of actual KOH still bonded together in that solution, which is almost nothing.  Mathematicians will tell you that dividing by a number close to zero will give you a number very close to approaching infinity, so the Kb value for KOH that you find in any table or whatever will never give you a number; it will probably just say "very large."
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
866213
KH2 is potassium hydride. This is not absorbable by plants. The phosphate is though (PO4).\n\nGo find yourself a good N-P-K fertilizer.
87390
There are several possible answers to this question, depending on what resources you have available to you.  If the solution you are measuring is known to contain only potassium carbonate dissoved in water, you can measure the potassium concentration by either atomic emission or atomic absorption methods and calculate to concentration of the potassium carbonate.  You could measure the carbonate concentration by titration or colormetric methods and calculate the potassium carbonate concentration.  Or you could simply evaporate a known quantity of the water to dryness, weigh the residue and calculate the concentration from those numbers.  \n\nIf you have other compounds in the solution, the last option, gravimetric, pretty much disappears.  The first two options are still practical, unless the other compounds contain potassium or carbonate.  Then, things get a little more complex.
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
192443
Write down the equation of reaction: K + Cl2 -> KCl (not balanced)\n\nBalance the equation of reaction: 2K + Cl2 -> 2KCl (balanced)\n\nFind the moles of Potassium: 5 g K / 39.10 g/mol K = 0.128 mol\n\nFind the moles of Chlorine gas: 7 g Cl2 / (2 x 35.45 g/mol) = 0.098 mol\n\nDetermine the stoichiometric ratios: \nCl2 -> 2 KCl\nK ->   KCl   \n\nFind the moles of product: \n\nFor potassium: 0.128 mol K x (1 mol KCl / 1 mol K) = 0.128 mol KCl\n\nFor chlorine:  0.098 mol Cl2 x (2 mol KCl / 1 mol Cl2) = 0.196 mol KCl\n\nSo potassium is the limiting reactant. \n\nMass of KCl that can be formed: 0.128 mol K x (1 mol KCl / 1 mol K) x (39.10 g/mol K +  35.45 g/mol Cl) =  9.54 g KCl
811861
KCl(s)  ---->  K+(aq)  +  Cl-(aq)
163249
Those are the correct names for potassiumchloride and potassiumphosphate
324860
pKa + pKb = pKw  so -log (2.1x10E-11) + pKb = 14\n\nSo pKb = 14- ( - log ( 2.1x10E-11 ) )\n\n= 14 - 10.68 = 3.32.... I swear I worked this exact question out when I was an undergrad....lazy students!  We didn't have the damned internet!  ( well we did but I didn't use it ----- go and learn the stuff!!!  Be a real chemist! )
1046324
What is your question kt?
141442
KE per mole = (3/2) RT where R = 8.3145 J/mole*K and T is temperature in kalvin, so:\n\nKE = (3/2) RTn where n is the number of moles of gas\n\nYou can solve PV=nRT for n and substitute to get KE = (3/2) PV where P is pressure and V is volume.
596849
2 - Potassium Chloride\n\nIt has, by far, the weakest covalency of all those listed, and therefore breaks apart most readily.
527802
N(Nitrogen), P(Phosphorus), K(Potassium)
847480
Q = m * Cp * DeltaT.\n\nPlug the numbers in.  Get Cp from your book.