What is the formula of Al+3 and Cl-1?
823404
AlCl3... the 3 is a subscript
844624
FeCl3   +    3NaOH  ----->    Fe(OH)3   +   3NaCl\n\nthats it
908581
C, Al(OH)3...\nWould you like an explanation? Okay, I'm supposed to be learning about this in Chemistry right now, so, I think this is how it works...\n\nIf you study the periodic table of elements, then you will find the number of valence (or outer) electrons each element has. Al has 3, O has 6, and H has 1. Before we start, it is important to know that a filled orbital has 8 valence electrons... okay? \nO and H combine, so they have a total of 7 valence electrons. You can imagine this by thinking of the letters OH surrounded by 7 dots in a circle, 2 pairs of dots on each side and 1 alone (a filled orbital with have four pairs opposite each other). The lone pair needs an electron, so it can borrow one from aluminum, which has 3 electrons. Aluminum loses its electron, so now it has 2, and the OH is now full, because it has 8 electrons. However, the aluminum has to get rid of its other two electrons, so another OH is brought in. The aluminum gives up an electron and the second OH is now full. Finally, the last OH pair is brought in and the aluminum gives away its last electron. So now, it is empty and the OH's are full. I guess they're balanced. Anyway, you are now left with 1 Al and 3 OH's, making Al(OH)3. Do you understand now? \n\nAl (3), and OH (7) (because O=6 & H=1, which equals 7)...\nAl - 1 = (2), so now OH = (8) (now it's full)\nAl - 1 = (1), so a 2nd OH (7) is brought in, making it OH (8)\nAl - 1 = (0), so a 3rd OH (7) is brought in, making it OH (8)\n\nIt took 1 Al and 3 OH's, so the new compound is Al(OH)3\nHope this helps!!!\nP.S. sorry for any confusion earlier when I tried to "draw" them. The computer changed the spacing so all you could see was a mess of stars and stuff. : )
889632
The Ag in the product can not appear out of no where. I assume the equation should have been written:\n\nAl + AgNO3  >  Al(NO3)3 + Ag\n\nSo . . . Al + 3AgNO3  >  Al(NO3)3 + 3Ag\n\nGet an "A".
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
244959
ALUMINUM ACETATE BASIC \n\nFormula: Al2O(C2H3O2)4.H2O
781803
--->H2O  +  H+  +  Cl-
878412
Na = +1 , O = -2\n\noxidation number Cl at NaClO = 0-1+2 = +1\n                              at NaCl = 0-1 = -1\nreduction can be defined as the reduction of oxidation number.\nfrom +1 to -1 (proved)
280181
ok so.  first   balance the equation. shouldnt be too hard. 2Cr  3Cl2   and     2CrCl3 \nchromium u dont have to worry about cuz itz excess. you have 125g of Cl2 though. so you have to find   out how many moles that is. by dividing by the molar mass. so once you find out the   number of moles reacting  and then  find out the according number of moles of CrCl3    that are formed. in this case  its the same number    so    then   once you know the number of moles. you    find the   mass by  multiplying by the molar mass. and there u go
127720
Assume you have 100 g of the compound (AlxCly)z where x,y,z are positive whole integers.  Based on the percentage you will have 20.23 g Al and 79.77 g Cl.  Now 20.23 g of Al is approximately 20.23g/26.98g=0.75 mole Al.  Similarly, 79.77g Cl is about 79.77g/35.45g=2.25 moles Cl.  The mole ratio now is x:y=0.75:2.25 or 1:3.  Therefore, your formula is (AlCl3)z.
550368
I think you have a notational error: did you mean Al2(SO4)3 for aluminum sulfate?\nIf that is the case, then proceed as follows: \nAl(OH)3 + H2SO4 -> Al2(SO4)3 +H2O\n1. First, balance the Aluminum atoms:\n\n2Al(OH)3 + H2SO4 -> Al2(SO4)3+H2O (2 Al)\n2. Next, balance the Sulfur:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+H2O (3 S)\n3. Next, balance Oxygen:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+6H2O (18 O)\n4. Hydrogen is already balanced (12 H)\nGenerally speaking, the strategy is to balance the least represented element first until you move to the most represented.(here it was aluminum, followed by sulfur). Hydrogen is usually the last one you balance.
746627
Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
956142
2Rb(s)+Cl2(g)--2RbCl(s)
516493
a) Toluene and Cl2 form p-chlorotoluene and HCl in presence of AlCl3\nb) nitrobenzene + Cl2 -(AlCl3)-> m-chloro-nitrobenzene\nc) chlorobenzene + Cl2 -(AlCl3)-> p-dichlorobenzene\nd) 1-(3-chlorophenyl)ethanone\ne) 3-chlorophenyl acetate\nf) methyl 3-chlorobenzoate\n\nOrganic Chem rocks!
625685
Aluminium has a valanly of 3 Oxgygen has 2\nSo\n4Al + 3O2   -> 2Al2O3
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
795171
0,240 M/L*2,60 L = 0,624 moles of Al(NO3)3\nNow, every molecule of this compound has 3 moles of NO3- ions, therefore\n0,624 moles of Al(NO3)3*(3 moles NO3-/1mol Al(NO3)3)=\n1,87 moles of NO3- ions