Chemistry Question - KCN?
917763
"Reagent Grade" means that a chemical meets standards established by the American Chemical Society with regard to purity.  R.G. chemicals are normally what one uses in the laboratory.  "Technical Grade" is the grade of chemical normally used in industry.  Such chemicals may contain impurities that can make them unsuitable for lab use.
748005
1500 mg KCL\n\nDo I get 20 points for anwsering the same question twice??
752218
2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O
944583
a) [O 2-]\nb) [ I - ]\nc) [Na+]\nd) [Al 3+]\ne) [Ni 2+]\nf) [Mg 2+]
433948
According to the MSDS for KCl,\nits solubility at 0 degrees C is 28.1 g/100 g of water
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
701443
1 - the reaction is reversed and multiplied by 4, so the enthalpy change is 11508 kJ\n\n2 - The reaction is a combination of the other three (I'll label them A, B and C, in the order you have them listed, for simplicity)\nThe original reaction = 2A - 2B -2C (the first one is just doubled, the next two are doubles and reversed), so 2*321.8 - 2*571.7 - 2*288.6 = -2542.2 kJ\n\n3 - This one is done similarly to question 2.  A - 0.5B + 1.5C\n180.8 - 0.5*(-906.2) + 1.5*(-483.6) = -91.5 kJ\n\nHope this helps
839352
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)
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
874059
I think that isolating potassium is the wrong approach.  Potassium metal reacts with atmospheric water to form KOH and hydrogen gas.  If you accidentally get water on the metal the reaction is so vigourous that the metal will catch fire, which will be dangerous.  \n\nTry a different approach.  Dissolve KCl in water and add a molar equivalence of Silver iodide, AgI, also dissolved in water..  AgCl is nearly insoluble in water (~0.9 mg/L at 10C)  and will precipitate out, leaving KI in solution.  Filter out the AgCl and evaporate the water, leaving solid KI.  \n\nThe two biggest problems starting with KNO3 are:  It will be nearly impossible to salt out the nitrate because of its high solubility in salt form, and two, many nitrate salts, particularlyof metals, are highly explosive and dangerous.  KCl is cheap and plentiful, so why not just buy some?
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
708591
Depends on your salt substitute, but the answer you're looking for is probably Cl- as your salt substitute probably contains KCl.
288242
The reaction should be written as follows,\n2 KClO3(s) --> 2 KCl(s) + 3 O2(g) H = -89.4 kJ\n\nYou either did not include the arrow or it is not displaying properly.\n\nPotassium Chlorate decomposes with heat to form Potassium Chloride and Oxygen gas, no where in this reaction is elemental Potassium involved.\n\nThe standard enthalpy of formation of any pure element in its natural state is zero. Oxygen is a diatomic molecule and thus, Oxygen gas (O2) has a enthalpy of formation of 0 kJ in this reaction, no matter how much of it you have, it is still zero.\n\nSince the Potassium Chloride is the only product left that has any non-zero enthalpy of formation, we know that the over all change in enthalpy is equal to the enthalpy of formation of KCl * the number of moles produced.\nIf the over all change in enthalpy is -89.4 kJ and 2 moles of KCl are produced, then the enthapy of formation of the KCl must be 44.7 kJ/mole less than the enthalpy of formation of the Potassium Chlorate.
543314
1.) Q = (nh3)^2/(n2*h2^3) = 0.16\n\nAs Q>Kp, reaction will move backward.\n\n2.) Kc = Kp*(RT)^(1+3-2) = 3.1*10^-4*(8.314*700)^2 = 3.55*10^11\n\n3.)\n\nGiven reaction is 2*(ii)+(i)\n\nSo, Kp = (8.3*10^-3)^2*3.1*10^-4=2.13*10^-8\n\nLet original P of HCN = 1\n\nso, partial pressure of C2N2 and H2 = p (assuming 2p of HCN decomposes at equilibrium)\n\nEqn is\n\np*p/(1-2p)^2 = 4*10^-4\n\nSo, p/(1-2p) = 2*10^-2\n\nSo, p = 0.019 atm.\n\nEnjoy.............
314792
NaNO3 + KCL = KNO3 + NaCl\n\nso you need the atomic weight of each element\ni.e.\nNa (23 g/mol)\nN (14 g/mol)\nO (16 g/mol) x3 = 48 g/mol\nor 85 g/mol for NaNO3\n\nK (39 g/mol)\nCl (35.5 g/mol)\nor 74.5 g/mol\n\nKNO3 is 39 + 14 + 48 = 101 g/mol\n\nyou need one mole of NaNO3 and one mole of KCL to make one mole of KNO3.  One ton is 1000 kg, or 1,000,000 g.\nSo 85 g/mole * 55 dollars/1,000,000g = 0.004675 dollars/mole\n74.5 g/mole * 28 dollars/1,000,000g = 0.002086 dollars/mole\nor added together, you spend 0.006761 dollars/mole to make KNO3\nNow, you can sell KNO3 for 185 dollars/1,000,000g, and\nKNO3 = 101 g/mol * 185 dollars/1,000,000g = .018685 $/mole,\n So ((0.018685-0.006761)/0.006761)*100% = 176% gross margin
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
771965
I can't tell you the name because i can't remember that stuff but the product will be K(NO3) PbI2\nif you're writing it in the long form it will be\n2KI + Pb(NO3)2 yeilds (arrow) 2K(NO3) + PbI2