What is the formula of coordination complex of Cobalt with ammonia as ligand?
438289
Co(NH3)4 would be more likely
804468
You need to look up the entropy for all compounds, multiply by the entropy by the stochiometric coefficients (e.g. 4 for NH3, 3 for O2). Then subtract the left side from the right side.\n\nVery simple actually, once you find the data.
910345
The formula would be Al2(CO3)3; the numbers are subscripts. \n\nLearn your charges for the polyatomic ions and for the other elements, it will really help.
839410
This compound is an amine.  You can identify it by the NH2 (amine group) at the end.  You did a fair job with the structure considering the limits of the text here.  I am sure you realize that two of your carbons do not have the correct number of bonds.
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2
913568
Zn(NO2)2; Co(HSO3)3; HgF; Al2(CO3)3; Cs2C2O4; Ba(ClO)2; SnBr2; Mg(HSO3)2.
935686
1. Look at the periodic table and figure out the formula mass of your compounds.  .005*formula mass of  Pb(NO3)2 and ditto for the 2nd one...\n\n2. 1/2 of 454g...
364266
Metal carbonyls are organometallic complexes of transition metals with carbon monoxide.
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.
541274
Bond order of CO is 3, bond order of CO+ is 2.5.  CO+ is paramagnetic.\n\nCO is (sigma)^2 (pi)^4, CO+ is (sigma)^2 (pi)^3
373250
C3H7NO2 \n\nwww.chemfinder.com is a useful resource.  \n\nI hope this helps :)
145111
The reaction formula is\n\nCH2COOH                           \nCH(OH)COOH (aq) + NaHCO3 (aq)----> \nCH2COOH                            \n\nCitric Acid\n\n       CH2COONa\n CH(OH)COONa    (aq)+ H20 (l) + CO2(g)\n       CH2COONa
825051
Covalent bonds, as in NH4+
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It's fairly easy don't worry.\n\nNo need for d level splitting diagrams to answer this since we're dealing with N and F:  use the spectrochemical series to make an assumption about the ligand.  The difference in t2g and eg levels (∆o) determines whether a complex is low or high spin.  If both ligands were the same, we would have to look at the oxidation state of the ligand in the complex.\n\nHalides <  Oxygen ligands  <  Nitrogen ligands < CN- ligands\n\n   ALWAYS HIGH SPIN    ---------------            ALWAYS LOW SPIN\n\nIf ∆o > P, then a complex is low spin.  IF ∆o < P the complex is high spin.
1177529
formula is C15H31COONa\n\nthere are single bonds between th 15 C atoms and also between C and H.\nyou can see the structure here www.kcpc.usyd.edu.au/discovery
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
721711
EAN Rule states that in forming a complex, ligands are added until the total number of e- in the central metal atom/ion becomes the same as the number of e- in the next inert gas.\n\ni.e. at.no. of the central metal atom+no. of e- lost in forming the complex+no. of e- gained by co-ordination should be equal to the no. of e- in the next inert gas.\n\nCr(CO)6\nat. no. of Cr+no. of e- lost in forming the ion+no. of e- gained (bcoz of CO).\n24+0+6*2=36 \nwhich is the at.no. of the next noble gas(Kr).\n(here the no. of e- lost in forming the ion is 0 bcoz Cr(CO)6 is a neutral complex & CO is also neutral in charge.Each CO group donates 2 e- , making a total of 12e- for 6 CO groups.\nHence it obeys EAN Rule.\n\nsimilarly you can work out for others also.\n\nwell the answer for your ques is Mn(CO)6 & V(CO)6.