How will this equilibrium be affected with the following changes?
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A) When temperature is increased the equilibrium shifts right because the left is is warm an unstable so by shifting right it will balance among both sides.\n\nB) Barium nitrate is 100% soluble in water and will therefore act like the products side of the equilibrium (the reactant is solid). The equilibrium shifts left to balance the amount of barium.\n\nC) When Ba(ClO3)2(s) is added the equilibrium shifts right for a similar reason as the first.
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the effects that it caused.
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If your question is about the effect of a change in Proportional gain on the step response of the Position Control of a DC motor, Then the Answer is that the change will result in a change in the speed of response to a step input.
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simply put: the higher the concentration the faster the reaction rate. in equilibrium the higher concentration shifts the reaction t o the other side.
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Le Chåtelier's principle - when a system at equilibrium is disturbed by application of a stress, it attains a new equilibrium position that minimized the stress.\n\nif a liquid substance is added, regardless if the substance is acidic or basic, the shift in equilibrium would be to the right\n\nBecause there would be a net movement of molecules from the liquid phase to the vapor phase, the number of liquid molecules has decreased.  The equilibrium has therefore shifted to the right
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It would definitely effect the field and the result would  be vector addition. If the source (here current) is supposed to generate a constant electric field then current gets effected but if there is no constraint then there is no reason for current to change.
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yes, it would be the same effect as changing shocks
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Check out the Bernoulli principle (try wikipedia).  That will explain all.
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ok.. first off the equilibrium expression is the product of concentration of products over the product  concentration of reactants, with any coefficients taken to the power of the of the coefficient For\n\nA+2B <=> C \n\nthe expression would be [C] /(over) [A] x [B](to the second)\n\n(that's C divided by A times  B-squared)\n\n____________________________________________________\n\nNow, in this particular problems the solutions are given to you. (Btw, in this solution i'm assuming your carbon concentration is carbon dioxide because there is no carbon the equation).\n\nYou should have [H2O] X [CO] /  [H2] X [CO2]\n\nSub in your contrations and you should get: .7864\n\nWARNING: I HAVEN'T DONE THIS IN A LITTLE WHILE... I'M A CHEM MAJOR, BUT I'M NOT GOING TO 100% GUARANTEE THAT
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omg i just had this test today too!!!!!\nu use hess's law to find the change--->write the equation, balance it \nand then use the moles (coefficients) to multiply the change is s for each element.\nhess's law says that the change in s (or h, or G, its usable for any of the thermo concepts) is equal to the sum of the entropy of the products minus the sum of the entropy of the reactants. its pretty easy after that--just plug in the values and there you go! \ngood luck! this was a hard test for me..so confusing.
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Yup. They are not simple...\nChemistry...\nhttp://www.chm.davidson.edu/ChemistryApplets/equilibria/EquilibriumConstant.html\nhttp://en.wikipedia.org/wiki/Chemical_equilibrium\nThey use the equilibrium constant expression (Kc) here...\nhttp://www.cofc.edu/genchemlab/equilibrium.htm
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Adding more chloride ion.  Adding a common ion shifts equilibrium to the left, due to the common ion effect.
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1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
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there are various factors which affects rate of reaction. like temperature,pressure and concentration of the reactants involved. and they all have different effects
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I predict the value will increase without bound. One factor in  any potential success is the way you integrate around the pole. There could be a strong singularity!
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if it is an ideal gas is definitely will\npV =nRT\np increase then T increase if V is stationary
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It depends when you change it - the rate fluctuates constantly. A good resource can be found at:\n\nhttp://www.xe.com/ucc/