balance this equation?
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.
1025264
pls specify ur question, didnt quite understand it\n\nIf there is "OR" u have to add and if it is "And" u have to multiply.
729087
1) this one is already balanced for you!!\n\n2)Count the number of Mg and O on the right side, and count the number of Mg and O on the right side.  On the left, you have 2 O and 1 Mg, and on the right you have 1 Mg and 1 O.  To make the oxygen on both sides equal, you need to double the MgO.  Now on the left, you have 2 O and 1 Mg, and on the right, you have 2Mg and 2 O.  Now you need to multiply the Mg by two so the Mg on both sides is equal.  and you're done!!!! \n2Mg + 02 --> 2MgO\n\n3) On the left, you have 1 Zn, 1H, and 1 Cl.  on the right you have 1 Zn, 2Cl and 2H.  the Zn are balanced and you just need to multply the HCl on the left to balance the whole equ.\nZn + 2HCl --->ZnCl2 + H2\n\n4) On the left, you have 1 P and 2O, and on the left you have 4 P and 10 O. If you multiply the P on the right by 4, and the O on the right by 10, you have 4P and 10 O on both sides\n4P + 5O2 ---->P4O10\n\n5)  On the left, you have 1 Fe 2 Cl, and pn the left, you have 1 Fe and 3 Cl.  The easiest way to balance this is multiply the Cl on the right by the amount of the left one and vice versa. when you do that, you have 1 Fe and 6Cl on the right, and you have 2Fe and 6Cl on the left. To finish the equation, multiply the Fe by two.\n2Fe + 3Cl2---->2FeCl3\n\nVoila!  You have balanced 5 equations!\n\ni hope this was helpful!
749096
Solve the inequality the same way you would any other equation. Just remember that when dividing or multiplying by a negative number you have to flip the inequality.
967368
Just treat all other letters as constants. You can multiply, divide, add and subtract both sides of an equation and maintain the equality.
138065
The equilibrium constant of any equation is [C]^c[D]^d/[A]^a[B]^b.\n\nthe equation is already balanced. But you have to know how many moles of each element there are in order to compute the problem
943054
Distribute the 2 to the two terms in the parentheses. Group like terms. Solve for X.
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
23536
is your balace wrong?\nI only figure below balance:\nMnO2+4HCl = MnCl4+2H2O
751075
Yes, you've got half the equation right. It should look like this:\nx = n+20
908510
It should be Fe2+ + 2OH- --> Fe(OH)2 since you not only need to balance the elements but you also need to balance the charges.
408554
yes. take out an x \nyou answer should be x(x+3)
703752
simply subtract 7 from both sides. This leave you with\n\nx<_3
851400
the first equ is balanced, b/c on the left side u have Pb-1, C-1 O-3 on the right side u have Pb-1 C-1 O-3 so since both side are equal it is balanced\n\nans for question 2 BeCl2 + 3O2------Be(ClO3)2\nBe-1                                  Be-1\nCl-2                                    Cl-2\nO-was 2 now 6                                     O-6
723776
Walk through these things slowly... First off, lets correct your first equation (?)\n\nLi + OH- + K2(2e-) <--> Li + 2e- + 2K + OH- \n\nThat doesn't make since... (is that your initial equation?)\n - Li are ballanced\n - OH- are ballanced\n - K2(2e-) goes to 2e- and 2K which is balanced\n\nWhat's the question again? I think there was a typo somewhere in your original expression. Are you going from LiOH to KOH? If so, your expression would be more like this...\n\nLiOH + K2(-2) <--> Li(+) + 2e- + 2KOH\n\nStart by balancing the OH and then the Li to get\n\n2LiOH + K2(-2) <--> 2Li(+) + 2e- + 2KOH\n\nAnd, there you go.\n - Li, K and OH are balanced
1116225
simply multiply it with 0.4
531205
2 + 2 = 4\n\nthats all I got