what is the Lewis structure of KMnO4?
147821
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines
243941
Answer was found in the Handbook of Chemistry and Physics\n\nKMn04 is only slightly solube and as you have aready found the\nsolubility constant.\n\nNaMnO4 is list in the hand book as Very Soluble both in cold and hot water, and did not give a constant. That means that a lot of NaMnO4 can be dissolve into water, somewhat like table salt, NaCl\n\nNaMnO4 is often found in liquid form at about 40% in water, and it is a dark purple liquid. you can get this data sheet \nby just type in Yahoo search "sodium permanganate solubility"\n\ngood luck\n\nDavid
87280
MBrO3-9H2O where M can be Na,K,Li, other +1 cations\n\nIf M is other than these you have to adjust subscripts
814611
CdSo4+NH4OH ---->Cd(OH)2 + (NH4)2SO4
763539
In the Lewis structure, dots left alone want to pair up with those of another atom.  If there are 5-6-7 dots all around, the atom will accept electrons to fill it last layer.  If there are 1-2-3 dots, the atom will want to give those electrons away.\n\nSo you look at your Lewis structures, and try to pair up alone dots from one giving atom to a receiving atom. \n\nFor your peroxyde, you would get H.   and :.O.: twice so one link between and H and a O, another link between the other O and the other H, and then the two O's link together.\n\nFor you chlorine, there are 7 dots all around each atom, so they will share their un-matched electron in a covalent bond.  \n\nFor the CH4O, their are 4 dots all around the carbon, so you would think that one H will bond to each (that's 4 H then) but to add the oxygen, you have to insert it between a H and the C (that the hydroxyde radical).  It is written CH3H usually to show the -OH polyatomic ion.\n\nFor the C2H4, the carbon has 4 dots all around it, and they will bond to each other.  Each will link to two hydrogen (to make up the four in the molecule), leaving two dots on each C, so this means that there will be one double bond between the two carbon atoms.\n\nFinally, the CH2O os trickyer.  That carbon, with its four dots, will have one H bond to each one dot, so there are two free electrons left.  That oxygen has six dots, two pairs and two free.  This means that each free electron of the O will link up to one of the free electrons on the C, which means that there is one double bond between the carbon and the oxygen.  You'll have H-C-H for the first part and C=O for the second part.\n\nIt's kinda hard to type these :-)  Hope this helps anyway
867434
Construct a Lewis structure for NH3 ( Use N as the central atom ):  first: Determine the total number of valence electrons in the molecule.  In this case there are 8.  Represent shared pairs of electrons between the bonded atoms.  There will be one line connecting each atom, with each line representing two electrons, accounding for six of the eight available electrons.  With the remaining two valence electrons, position them on the fourth unbonded side, satisfying the octet rule for Nitrogen.\n\nThe above described Lewis structure gives you the molecular structureTrigonal Pyramidal ( three bonding pairs and one non bonding pair)\n\nRules for predicting molecular shape by VSPER theory:\n\n1.  Determine Lewis structure\n2.  Determine the number of electron pairs (bonded and non bonded) surrounding a central atom.\n3.  Use VSEPR theory to decide what symmetrical three dimensional shape will accommodate all bonded and non bonded electron pairs.\n4.  Classify the shape of the molecule by considering the arrangement of the joined atoms only, ignoring the nonbonded electron pairs.  That is why, although the NH3 molecule  has four pairs of electrons around the central atom, ammonia's shape is trigonal pyramidal.
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.
200262
mono = 1\ndi = 2\ntri = 3\ntetra = 4\npent = 5\nhex = 6\nhept = 7\noct = 8\nnon = 9\n\nnonmetals:\nif there is no prefix for the first element, it is assumes to be one.\nTHe ening of the anion (the second one listed) is changed to -ide.\n\ncarbon - 1 \noxygen - 2\nCO2  <---The number is a subscript, but you cant do that on here\n\nPhosphorous - 1\nIodine - 3\nPI3\n\nHydrogen - 2\nOxygen - 1  (it should be monoxide)\nH2O\n\nHope that helps you.  Don't feel bad; you can't know what you haven't learned.
555626
The Lewis dot structure would be that N in the middle with branches off to oxygen and 2 Fs and then one single electron will be on nitrogen and it is a radical I think. It is very reactive because of this one electron. The electron geometry will be something with sp3 orbital, I am not sure since it only has one electron. I would check up on that. Hope that helps
117753
Okay you have a choice between Potassium oxalate (K2C2O4 ) and Sodium Iodide (NaI). \n\nPotassium oxalate is highly soluble (30g/100g) and sodium iodide is even more soluble(184 g/100 ml).\n\nWithout knowing the reactants it is almost impossible to predict a product, especially with two choices. Given those two products, I would expect it to be K2C2O4, but that is just a guess. What was in the solution?
370118
if it forms a 2- ion it is from the Chalcogen (group V1(A) or 16) family (Oxygen, Sulphur, ...).  The two bonds between two Oxygen atoms are Covalent.  It has 6 valence electrons and takes two to complete the valence shell (noble configuration).\n\nThe reaction of X with Sodium (Na) will require twice as many Na atoms as Sodium only has 1 valent electron to give (to make the Na+ ion)\n\nSo\n\n2Na + X -> Na2X which in a polar solvent (e.g. water) will become 2Na+ +X2- \n\nFYI, the Halogens are Fluorine, Chlorine etc and only have one single space left to take an electron to reach Noble configuration hence they form singly charged anions F-, Cl-...
842680
2 KBrO3 + 3 N2H4 -> 2 KBr + 3 N2 + 6 H2O\n\nYou can check by finding out the total number of atoms of both sides:\n 2 Potassium\n 2 Bromine\n 6 Oxygen\n 6 Nitrogen\n12 Hydrogen
613727
try:\nhttp://lb.chemie.uni-hamburg.de/static/data/50_zzk46n8n.html
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
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)
753744
ANSWER:  MnCl4\n\nGiven the info above, you can use the Ideal gas law to find the mol of H2 used.\n\nn = PV/RT   so;\n\n(0.951 atm)(3.22L)/(0.0821atm*L/mol*K) = 0.10mol H2\n\nUsing the mass of Mn we can find the mol of Mn:\n\n2.747g(1mol Mn/54.938g) = 0.05 mol of Mn\n\nAs you can see, the mol ratio of Mn to H2 is 1:2.  \n\nFrom this you can write the balance equation between Mn and HCl to form Manganese compound.  Giving 1 to 2 ratio between Mn and H2:\n\nMn + 4HCl ---> MnCl4 + 2H2\n\nAs you can see, the manganese compound formed is MnCl4, Manganese(IV) chloride.