is delta G for the dissolving of AgNo3(s) psitive,negative or zero. why?
344369
If it were not negative it would not take place.  That is the point of delta G - it is a measure of free energy based on a combination of the heat released (delta H) and the entropy (randomness) - delta S.  If delta G is not negative for a given process under a given set of conditions, then that process will not occur.  If the process does occur, then delta G is negative.  Now - be careful - just because delta G is negative does not mean a process will neccesarily occur - or at least occur at an appreciable rate - in amny cases you need activation energy to "get the process over the hump" to make things happen.  Example - matches - they ignite - that is a negative delta G process - but it doesn't happen spontaneously - at least not at room temperature - you need to activate the process by heat or friction.
744648
Use Gibbs free energy equation\n\nDeltaG=DeltaH - TDeltaS\n\nYou want Delta G to be negative so if Delta H positive then you need a greater TDeltaS \nDelta S is -25 so in order for the reaction to be spontaneous the T has to be less than zero
851188
when a piece of Ag is dropped in FeSO4 sol. no reaction will take place. it is because Ag is less reactive than Fe so it cannot displace Fe in FeSO4
307574
No, it's Ag2S. The 2 is a subscript, denoting 2 Silver atoms to 1 Sulphur atom. Silver has a valency of 1+, while Sulpur has a valency of 2-, hence 2 Silver atoms are needed to balance with 1 Sulphur atom to form a compound. Besides, the equation would make sense then. 2 Silver atoms and 1 Sulphur atom on both sides of the equation, the equation is now balanced. 10 marks please, thank you!
533621
(delta)H(rxn) = sum of heat of formation of products - sum of heat of formation of reactants.\n\na.  Hf Br2(g) = 30.91 kJ/mol, Hf Br2(l) = 0, so (delta)H = -30.91 kJ/mol\n\nb.  Hf CaCO3(s) = -1207 kJ/mol, Hf CaO(s) = -635.5 kJ/mol, Hf CO2(g) = -393.5 kJ/mol, so (delta)H = -635.5 - 393.5 + 1207 = +178 kJ/mol\n\nc.  Hf NO(g) = 90.25 kJ/mol, Hf O2(g) = 0, Hf NO2(g) = 33.2 kJ/mol, so (delta)H = 2 * 33.2 - 2 * 90.25 = -114.1 kJ/mol
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
876161
You're right - as written it's impossible to balance.  But if you write\n\na(Ag2CO3) --> b(Ag2O) + c(CO2) + d(O2) you can balance it with the values \na=1, b=1, c=1, d=0.    Meaning that the reaction doesn't generate any oxygen.  The carbonate falls apart into the oxide and gaseous CO2.
892172
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
921926
mg + ion\no -ion\nk +ion\ns -ion\nwhen they are 'added' together (MgO and K^2S) the charges are balanced, that means that the net charge is 0.
341402
Specific heat does not apply to reactions, only actual chemicals.  You are probably thinking of the heat of reaction, or enthalpy.  The change in enthalpy of a reaction (written as delta H, delta as the Greek letter that looks like a little triangle) flips its sign when the reaction is run in the reverse direction.  So a positive delta H value becomes negative when the reaction is run in reverse direction, and a negative delta H value becomes positive.
816266
At constant temperature and pressure, a reaction will occur spontaneously if the Gibbs free energy change of the reaction is less than zero.\n\nThe Gibbs free energy change, delta-G, is given by:\n\ndelta-G = delta-H - T*delta-S\n\nwhere T is the absolute temperature, \n\ndelta-H is the enthalpy change for the reaction at the temperature and pressure of interest, and \n\ndelta-S is the entropy change for the reaction at the temperature and pressure of interest.\n\nWe are given the enthalpy and entropy changes for the reaction of interest in this question.  I will assume that the enthalpy change and entropy change for this reaction are independent of temperature. Note also that the values of both of these thermodynamic parameters are given in "per mole" units in this question.  I will assume these values are normalized to one mole of one of the reactants, and not one mole of the product.\n\nPlugging in the enthalpy and entropy values, and setting delta-G to be less than zero:\n\n0 > delta-G = 180500 J/mol - T*24.8 J/(mol*K)\n\nT  > (180500 J/mol)/(24.8 J/(mol*K))\n\nT > 7278 K\n\nThis is an extremely high temperature, and the assumption that we made above that the enthalpy and entropy changes were independent of temperature is probably not very good given the large temperature extrapolation being made here.
390546
I'll give you a hint.\n\nDelta G = Delta H - T Delta S\n\nUsing that formula, you can get 1\n\n(B) you can look up in any physical chemistry text:\n\nG(2) - G(1) = a formula I don't have available
197904
based on the net inoic equation:\n\n\nCu2+(aq)---> Cu(s)\n\nAg+(aq)---> Ag(s) --- zero grams\n\nexcess Cu(s) or Cu(aq)\n\nif, Cu(s) then it is  3.78 grams of Ag
51552
AgNO3 + NaCl -----> AgCl + NaNO3\n\ntotal ionic equation\n\nAg+  +  NO3-  +  Na+  +  Cl-\n\n                      ----->\n\n                            AgCl  +  Na+  +  NO3-\n\nnet ionic equation\n\n    Ag+  +  Cl-  ----->  AgCl
896452
Not sure the level with which you are learning, so feel free to ignore the thermodynamics explanation.  If you just want to answer the question, select D.\n\nLike most test questions, there is not enough conditions specified and truly, we could choose several anwers, but only after making assumptions that less likely to be in the testers mind.  I will assume a high school or intro level college course because above that this should be challenged as too vague.\n\nUse logic to realize that answer choice C is not true. Evaluating the others we realize that A means only heat is given off when bonds form and B means heat is absorbed or added during the formation of bonds.  D means conditions exist in which either situation is possible and E means that final product is in a state that has a higher value for entropy.  This only holds true when both Temp and pressure of a system is held constant, but we will eliminate later as well.\n\nBasically a bond can occur when the system goes from a high energy state to a lower energy state.  Formation of bonds is essentially a chemical reaction.  The Gibbs Free Energy equation:\n\n (delta)G=(delta)H-T(delta)S\n\nT=temp of system in Kelvin\n(delta)H is the heat energy change\n(delta)S is the change in entropy (disorder)\n\ndefines whether a reaction will occur spontaneously or not.  (deltaG) must be negative for this to be true. So we will assume \nusing this equation and recognizing that (delta)H tells us exo or endothermic if it is positive or negative , examination of the possibilities of (delta)H and (delta)S,\n\nIf (delta)H is negative and (delta)S is positive the equation will result in a negative value for (delta)G which means a favorible condition exists for bonding to occur.  Therefore, the condition of (delta)H negative (energy leaves as heat - exothermic) is true.  We can thus eliminate answer choice B.\n\nIf the opposite is true and (delta)H is positive and (delta)S is negative, the result is (delta)G that is positive and therefore not spontaneous.  no reaction or bonding occurs.\n\nIf both (delta)H is positive and (delta)S is positive, only a large enough (delta)S will overcome the positive (delta)H in the equation.  So if both are positive and T(delta)S is greater than (delta)H the bonding will occur.  Therefore, when (delta)H is positive (endothermic) the reaction may occur as well.  We can thus eliminate choice A.\n\nWe have shown that C is a true statement because conditions can be met in which the reaction is both exo or endothermic.  However, please note that in both conditions (delta)S is also positive, indicating an increase in entropy.  For the most part any system will always trend toward an insrease in entropy.  \n\nThe question then arises, what happens when (delta)S is negative (decreased entropy) and (delta)H is negative.  For most situations T(delta)S will be large enough to create a positive (delta)G from our equation.  In very rare cases will (delta)H - T(delta)S result in a negative answer.  However, the equation shows this as a possibility when (delta)G= large negative (delta)H - T x small negative (delta)S => large negative # minus a smaller negative => larger negative + smaller # = negative.  (Hope you follow my train of thought).  Therefore in the right situation - (delta) can be negative and bonding/reaction can occur when entropy decreases.  Eliminate E and C is the correct answer.\n\nCareful what you ask for.  Wikipedia site looks pretty informative but very detailed and you have to use a few different pages.
670034
Ag+ will precipitate in HCL solution as AgCl, but as for Ag(NO3), i am not so sure.
133194
This usually happens in solution, the Chlorine is a chloride ion.\n\nAgNO3(aq) + Cl-(aq) --> AgCl(s) + NO3-(aq)