What is the formula for 2-napthoin?
603629
1,2-Bis-(2-naphthyl)-2-hydroxy-ethanone (2-Naphthoin)\n\n.................O\n..................||\n(2-naph)-C-CHOH-(2-naph)\n\nwhere (2-naph) represents the 2-naphthyl group [C10H7-]\n\n(ignore the dots in the structure above -- they are used to hold the spaces that Yahoo otherwise deletes)\n\nTotal empirical formula would be: C12H16O2
836780
Its 5=2n-6 \n\nwell done!
634218
There really is not a set formula, as the -ite anions don't have a set formula themselves (PO3 2- for phosphite, NO2- for nitrite, SO3 2- for sulfite, Cl02- for chlorite, etc.)
267313
The only think that would make sense to me would be\n\n4NO3 + 2H2O = 4HNO3 + 02\n\nalthough i am not really sure\n\nGood question though!!
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
859821
2.2 gram\nu can calculate it by the molarity formulae
778796
bcd\n\n\n\nSolution:\nCaCN2 + 3H2O ---> 2NH3 + CaCO3
869137
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba (PO4)2 +  2Ba
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2
921488
2ClO4(-) --->  2ClO4(-)\n2H(+) + 2OH(-) ---> 2H2O\n\nwhere sign in brackets are superscripts.
808775
Na2C2O4\n Na +1, C +3, O -2
756921
the answer is\n2Al + N2 ----> 2AlN
743363
1. Get the number of moles of carminic acid.\nsince 1 mole of NaOH neutralizes 1 mole carminic acid, number of moles NaOH equals number of moles carminic acid:  \n0.01802L * 0.0406mol/L = 7.3 x 10^-4 mol\n\n2. Determine the molar mass of carminic acid.\nSince you are given the weight of carminic acid, you can get the molar mass by simply dividing it with is number of moles from #1.\n0.3602g / 7.3 x 10^-4 mol = 492.34 g/mol\n\n3. Calculate for the molecular formula.\na) Assume 100g carminic acid such that weights of C, H, and O are as follows: C = 53.66g; H = 4.09g; and the remaining \nO = 42.25g\n\nb) Get the number of moles of each element by dividing the weights by atomic mass. \nC => 53.66g/ 12g/mol = 4.47mol\nH => 4.09g/ 1g/mol = 4.09mol\nO => 42.25g/ 16g/mol = 2.64mol\n\nThis is your empirical formula: C4.47H4.09O2.64\n\nc) To get the molecular formula, the molecular mass must be equal to 492.34 g/mol. In this case, you have to multiply the empirical formula by ~5 to get C22H20O13. To check,\nC = 22* 12g/mol = 264g/mol\nH = 20* 1g/mol = 20g/mol\nO = 13* 16g/mol = 208g/mol\nwhen you add all up, you get 492g/mol.\n\nSo, molecular formula for carminic acid is C22H20O13.
904268
just add 3  zeros or multiply by 1000 (which is the same as adding 2 zeros o_O)\n\nexample: 78g=78000mg
814611
CdSo4+NH4OH ---->Cd(OH)2 + (NH4)2SO4
832115
N2H4 + H2O -> N2H5+ + OH-\n\nKb = c[N2H5+] * c[OH-] / c[N2H4]\n\nAssuming that c[N2H5+] = c[OH-] and that c[N2H4] >> c[OH-] (so that we can use the starting concentration for c[N2H4]), we get approximately\n\nc[OH-] = sqrt( Kb * c[N2H4])\n\nor logarithmized\n\npOH = 1/2 ( pKb + 10^-(c[N2H4])\n\npH= 14 - pOH\n\n\ncalculator time.