What is the expected empirical formula for a compound made from gallium and iodine?
924763
Since Ga is under Boron and Aluminum, which both have a valence of +3, you would expect Ga to behave the same way, losing electrons from the p and s orbitals (but not the d).\n\nSo b) GaI3 is correct
956142
2Rb(s)+Cl2(g)--2RbCl(s)
795424
First, use the ideal gas law PV=nRT to solve for n, the number of moles.  You are given P,V, and T.  R is the gas constant 62.364 mmHg*L/(mol*K).  Now you have the number of moles of the unknown and you were given its mass.  Thus, you can find the molar mass of your unknown.  This will help to determine your molecular formula when you compare this value to that of the empirical formula.\n\nTo determine the empirical formula: Assume a 100g sample.  Then you'll have 65.45 g C, 5.49 g H, and 29.06 g O.  Determine the molar ratios of each compound and divide by the least common multiple.  In this case, you get a ratio of C:H:O of roughly 3:3:1.  So your empirical formula is C3H3O.  Compare the molar mass of the empirical formula (55 g/mol) with the molar mass calculated above.  Multiply the components of the empirical formula by a scalar as necessary to match the molar mass of the unknown.
24431
3225..............x\nZn   +  2HCl -> ZnCl2 + H2\n65.4............136.4\n\nx=3225*136.4/65.4\nx= 6 726.14679 grams
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
91712
83.93 grams of GaCl3 are produced.\nHow?\nWell, note the following:\n35.453 g Cl/mol Cl\n70.91 g cl2/mol Cl2\n69.723 g Ga/mol Ga\n\n1 mole of GaCl3 = 3(35.453 g Cl) + 1(69.723 g Ga) = \n176.082 g GaCl3\n\n(50.7 g Cl2)*(1 mol Cl2 / 70.91 g Cl2)*(2 mol Cl/1 mol Cl2)*(1 mol GaCl3 / 3 mol Cl)*(176.082 g GaCl3 / 1 mol GaCl3) = \n83.93 g GaCl3\n\nThis means that for every 3 moles of Cl used, there will be 1 mole of GaCl3 produced.  It also means that for every mole of Cl2 used, 2 moles of Cl are going into the product.  Therefore for every 3/2 moles of Cl2 used, 1 mole of GaCl3 is produced.
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
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...
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
947888
The atomic number of gallium is 31. Therefore, there is 31 protons and 31 electrons. However, The mass number of gallium can be varied because gallium contains isotpes. Each isotope has the same atomic number but different mass number. The most abundance isotope of gallium has a mass number of 70. So there is (70 - 31) = 39 neutrons.
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6
1093944
i) add NaNH2(1 eq.)  then CH3I(1 eq.) (I=iodine)then NaNH2(1 eq.)  then C2H5I then add hydrogen in presence of palladium and BaSO4.\nii)add NaNH2(1 eq.) then C3H7I(1 eq.)  then add NaNH2(1 eq.)  then again C3H7I(1 eq.) then add ZnBr2 or MgBr2.
743385
Cl2 + 2 KI → I2 + 2 KCl
759497
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.432 g x (12 / 44) = 2.027 g\nhydrogen: 4.057 g x (2 / 18) = 0.451 g\noxygen: 3.383 - 2.027 - .451 = 0.905 g\n\n2) convert to moles.\ncarbon: 2.027 / 12 = 0.169 mol\nhydrogen: 0.451 / 1 = 0.451 mol\noxygen: 0.905 / 16 = 0.0566 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.169 / 0.0566 = 3\nhydrogen: 0.451  / 0.0566 = 8\noxygen: 0.0566 / 0.0566 = 1\n\nThe empirical formula of the substance is (C3H8O)n\nThe atomic masses add up to 60, which is also the molar mass of the compound. Therefore n=1 and we find C3H8O
829175
AB + CD ---> AD + CB\nlogic doesn't work here as Si doesn't react/bond with Ca or Phosphorus.and Ca(PO) should be Ca3(PO4)2\nand SiO should be in the form of SiO2. \nso unless you want hypothetical compounds SiO2 won't form with either of the two other reactants.\nbased on the forumlae from an SI data book, the above reaction would not balance.\ngood luck solving this one. :-)
852476
Ga= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1\nGe= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2\nAs= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3\nSe= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4\nBr= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5\nKr= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
847953
If you know the mass ratio of the components of a compound you can determine the empirical formula.  What you have to do is convert the mass ratio into a percentage of the compound.  Then assuming you have a 100 gram sample the percentages would be the grams of each element.\n\nThen divide the grams of each element by each element's atomic mass.  This gives you the moles.\n\nThen divide the moles of each element by the smallest value to get an integer value.  If one or more of the values is not an integer, multiply by the smallest number you can to get whole numbers.  Then the numbers will be the number of atoms of each element in the compound.\n\nYou determine the mass ratio by quantative analysis.