Oxalic acid's comp. is 26.7%C,2.2%H,71.1%O. its molecular weight is 90amu. What is its molecular formula?
961793
Call it (CxHyOz)\nMass of C = 12x\nMass of H = 2y\nMass of O = 16z\n-----------------------\nIts molecular weight M = 12x + 2y + 16z = 90\n\nPercentage of C: (12x / M) *100% = 26.7% <-> 12x = 90 * 26.7 / 100 <-> 12x = 24 <-> x = 2\n\nYou can do the same with H and O to find y = 2 and z = 4.\n\nThe formula is, hence, C2H2O4.
933864
to what?  Are you looking for the molecular weight of the monoprotic acid?  If so, remember M (molarity) is in moles/lter) and your colume of NaOH is in mL so you'ww want to convert that to liters.\n\n\n.01524 liters x .1578 moles/liter = 0.002405 moles\n\nso therefore you have 3.567  g / 0.002405 moles = 1483 gram/mole\n\nuse dimensional analysis and keep track pf the units to find the answer
151307
N-(4-hydroxyphenyl)acetamide\n\nC8H9NO2
174643
The formula in the above answer is incorrect since it omits the aromatic ring and the 9 subsequent repeating units (it's not 9 repeating -OH groups, that makes no sense).\n\nThe empirical formula by my calculations seems to be: C33 H60 O10
350659
Why do people answer complete and total bullcrap answers?\n\n\nThe only thing they were right about was it is a weak acid, and even that bears no relavance to its P.H only you use a diferent formula in calculating it\n\n\nthe question is incomplete, you need to know the molarity of the acid and then you get the square root of it disociation constant (Ka) multiplied by its molarity\n\nand then you -log(base10)the answer you got from the first part and that is the P.H\n\nin short  Sq root(Ka*molarity acid)= H+ concentration\n\nP.H= -logH+
543389
I will explain it by an example.\nA compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound?\n\nGet the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).\n\n(.7546) (100 g) = 75.46 g C \n(.0443) (100 g) = 4.43 g H\n\n(.2010) (100 g) = 20.10 g O\n\n\n\nConvert the mass of each element to moles.\n\n(75.46 g C) (1 mol/ 12.00 g C) = 6.289 mol C \n(4.43 g H) (1 mol/ 1.008 g H) = 4.39 mol H\n\n(20.10 g O) (1 mol/ 16.00 g O) = 1.256 mol O\n\n\n\nFind the ratio of the moles of each element.\n\n(1.256 mol O)/ (1.256) = 1 mol O \n(6.289 mol C)/ (1.256) = 5.007 mol C\n\n(4.39 mol H)/ (1.256) = 3.50 mol H\n\n\n\nUse the mole ratio to write the empirical fomula. \n\nMultiplying the mole ratios by two to get whole number, the empirical formula becomes: \nC10 H7 O2\n\n\n\nFind the mass of the empirical unit.\n\n10(12.00) + 7(1.008) + 2(16.00) = 159.06 g/mol\n\n\nFigure out how many empirical units are in a molecular unit.\n\n(318.31 g/mol) / (159.06 g/mol) = 2.001 empirical units per molecular unit\n\n\nWrite the molecular formula.\n\nSince there are two empirical units in a molecular unit, the molecular formula is: \nC20 H14 O4\n\nI think,now u got an idea about the compound's formula
957775
Ca(OH)2 + CO2 --> C(OH) + CaO\n\nIs unbalanced..
969093
H2 = 2g/mol   2.05g = 1.025mol\nCO = 28g/mol    15.85g = .57mol\n\n2H2 + CO --> CH3OH\n\nmust be reduced to \nH2 + (1/2)CO --> (1/2)CH3OH\n\n1.025mol + (1/2).57mol --> the quantity of the limiting reactant which is .285mol CH3OH which at 32g/mol = 9.12g
777607
Since they are both strong acids and bases you can assume: \n\nMaVa = MbVb\n\n(25/1000) * (1.2) = Mols NaOH\n\nconvert Mols to grams and that's your answer
1149538
formula of ethylene glycol is C2H6O2\n                            ie CH2OH-CH2OH\napproximate atomic masses of C=12\n                                                   H=1\n                                                   O=16\n   therefore,\n2*12+6*1+2*16\n= 24+6+32\n=62g/mol
797533
Providing that the H3C6H5O7 is citric acid, that balanced equation is:\n\nH3C6H5O7 (s) + 3 NaOH (aq) --> Na3C6H5O7 (aq) + 3 H2O (l)\n\nThe net ionic equation is:\n\n3 H+ (aq) + 3 (OH-) (aq) --> 3 H2O (l)
777814
Q1)\nCa3 (PO4)2  (MW)molecular weight = 3*40 + 2 * (95) = 310 the oxygen weight is 16*4 = 64 .so % of O is (64/310) *100 = 20.6%\n\nQ2)\nSO4 (NH4)2 ---> MW = 32+4*16+ 2*18 =132 Mass of sulfur 32 \nhence  % S = 100*32/132 =24.2\n\nQ3) \n\nFe2 (SO4)3 ----> MW = 56*2+96*3=400 \n\n% of iron 100*112/400 = 28%
247135
The formula for water is H2O i.e. there are 2 atoms of hydrogen and one atom of oxygen. Its molecular weight is 18 g/mol .
384774
CH3CH2OH\n\nC ~ 12\nO ~ 16\nH ~ 1\n\n2*C + 1*O + 6 * H = 2*12 + 1*16 + 6*1 = 46 g/mol
757438
CaC2 + 2 H2O ---> Ca(OH)2 + HCCH
979624
Molar mass of ( CaCl2x2H2O)= mass of (Ca) + 2mass of (Cl)  + 4mass of (H) + 2mass of (O)\n\n\nMolar mass of ( CaCl2x2H2O)= (40g) + 2x(35.5g) + 4x(1g) + 2x(16g) \n answer \nb) 147 g/mole
620217
HOCl <==========>  H+  + OCl-\n\nthe pKa of hypochlorous acid, HClO, is 7.4 at 25°C, so:\n\npKa = -log [Ka] = 7.4, therefore\n\nKa = 3.98 x 10^-8\n\nKa = [H+]*[OCl-]/[HOCl] = 3.98 x 10^-8\n\nfrom your problem, you state that you have 1.0M HOCl, and since the only source for ions is the HOCl dissociating;\n\n[H+] = [OCl-] = y, and [HOCl] = 1.0-y\n\nsubstituting this back into the Ka equation, we get:\n\n3.98 x 10^-8 = y^2/(1.0-y)\n\nwhich converts into the quadratic equation:\n\n0 = y^2 + (3.98x10^-8)y - 3.98x10^-8\n\nI leave it to you to solve the quadratic equation to get y = [H+], and then calculate the pH of the starting solution