In Einsten's famous equation what applies? Like does it have to do with chemical reactions, collisons between
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In short, yes. But E=mc^2, is so much deeper than chemical reactions. Actually, it relates mass to energy. It lead to the concept of nuclear fission. It says that if you have a small amount mass, say 30 AMUs, and you multiply that by the speed of light squared, you will get a TREMENDOUS AMOUNT OF ENERGY. It was derived in Einstein's paper on relativity in 1905.
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Molecules and atoms interact with one onother during a chemical reaction. If atoms are indivisible then they cannot be destroyed during a chemical reaction. If atoms cannot be destroyed then the mass of reactants must equal the mass of the products in a chemical reaction. e.g.,\n\nReactants -------> Products\n\nMass of Reactants = Mass of Products
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The equation is meant to record the results of an experiment.  The reactants (stuff you started with) are written to the left and the products (the stuff you end up with) is written to the right.\nTo balance an equation, only the coefficients (the numbers in front of the chemicals) may be changed.\nIf you change the symbols, you change the type of atoms used.\nIf you change the formulas or subscripts, you change the chemicals (the molecules) used in the experiment.
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H2S---> 2 H^+ + S^2-\nK(eq) = \n[2H^+]^2 [S^2-]\n ---------------\n     [H2S]\n\nNH4+----> NH3  + H^+\nK(eq) =\n [H^+][NH3]\n       ------------\n         [NH4+]\n\nC6H5COOH----> C6H5COO-  + H+\n\nK(eq) = \n[H^+][C6H5COO-]\n        ---------------\n         [C6H5COOH]
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To say its an equation involving the relation between energy possesed by a mass
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The number of atoms/ions of an element on either side of the equation.\nThe total charge on either side of the equation.\nThe elements present are the same on either side of the equation.
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Yeah, it's E=mc^2 and it relates energy and mass.  E is energy, m is mass, c is the speed of light (celeritas).  The equation could come useful in physics, like atomic physics.\n\nWhat did you wanted to know?
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the actual equation is:  (p + b·n^2/V^2)(V - a·n) = nRT\n\na:- every mol of particles of the gas has a tiny finite volume of a\n\nb:-modification is due to the particles sticking to each other, a phenomenon called cohesion\n\nR- rydberg's constant
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Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
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A balanced equation is one where the number of atoms of each element are the same on both sides of the equation. This means it will follow the Law of Conservation of Mass.
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the equation is as simple as the elegant and simple universe.the equation means that any type of energy can be converted into mass and vice versa. it means that at speed of light the enrgy becomes infinite(to accelerate it actually)and thus mass becomes infinite(special relativity)provided the generation of light in both the cases. the actual equation is e(kinetic energy)=mc^2/underroot(1-v^2/c^2).the e=mc^2 equation is still applicable for objects moving or stationary meaning that both the equations are true.
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2CO(g) + O2(g) --> 2CO2(g)
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Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
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Coulombs law.\nWhen the distance is made 3 times longer (for instance),\nthe force between the charges becomes 3^2=9 times smaller.
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An example of a compound form when carboxylic acid and alcohol react.\n\nAn equation for such a reacton is\n\n R-COOH + R'-OH = R-COO-R' + H2O \n\nthe -COOH group and the -OH group lose a water and become a -COO- linkage: \n\n\nwhere R and R' represent organic groups.
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The index-of-reaction equation (i.r.e.): dF/dtZ = kZ(l - βF)m is compared with a generalized Prout-Tompkins equation (g.P.T.e.): dF/dt = kn(l - βF)y With the boundary condition F = 0, t = 0 the equations are equivalent for z = 2 or m = 0 or β ≤ 1 and the i.r.e, is closely approximated by the g.P.T.e. for other values of the parameters by use of the following approximation ∫(1- F)-m dF ≈ AF(1-BF)5 The parameter s is a function of m. Both A and B are close to unity for 1/3 < m < 3. The i.r.e. is recommended as being more useful than the g.P.T.e. because the i.r.e. has a better theoretical basis and because it is more readily integrated and fitted to experimental data.
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Neutralization Reaction:\nHC2H3O2(aq) + KOH(aq) -->H2O(l) + KC2H3O2(aq)\nThe equation is already balanced.\n\na. False KC2H3O2(aq) and H2O are products.\nb. True\nc. False, some ions will cancel out but some will remain.\n\nThe ionic equation is as follows:\n(H^+)(aq) + (C2H3O2^-)(aq) + (K^+)(aq) + (OH^-)(aq) -->H2O(l) + (K^+)(aq) + (C2H3O2^-)(aq)\nRepeated ions are crossed off from both sides and the net ionic equation is:\n(H^+)(aq) + (OH^-)(aq) -->H2O(l)\n\nd. True\ne. False\n\nNote: the + and - signs were written with ^ to indicate that they were raised up (like a power).  Sorry if this notation is confusing.  Good luck and I hope that this helps!