Why is it fair to assume that the hydrochloric acid solution has the same density and specific heat capacity?
966389
Because it is quite a mild solution. There's plenty more water than acid, therefore you may assume that the properties of the water haven't changed much.
842422
Yes, it is true that the density of a solution is different of solvant but tjis difference is small ; So , i propose to dissove 40g of amyl acetate in a liter of water if that acetate is a solid. If that acetate is aliquid the total volume will be 1.04 l and you must use 47 cc tp water
813561
The molecular mass of NaOH is 40.00 gm/mol,  so 8 gm of NaOH contains 8 gm/40.00 gm.mol = 0.20 moles of NaOH.\n\nThe molecular mass of HCl is 36.46 gm/mol, so 8 gm of HCl contains 8gm/36.46 gm/mol = 0.22 moles of HCl.\n\nIf we react these amounts of NaOH and HCl, the NaOH will be the limiting reactant, and we will produce 0.20 moles each of NaCl and H2O.  This reaction will liberate \n\n0.20 mol * 56 kJ/mol = 11.2 kJ of energy as heat\n\nThe heat capacity of the final solution is given by the mass of the final solution (which, by conservation of mass, is equal to the mass of the initial solutions) multiplied by the specific heat capacity of the final solution.  With the assumptions given in the question, the heat capacity of the final solution is:\n\n(58 gm + 258 gm) * 4.184 J/(gm*degC) = 1322 J/degC.\n\n(Note that we don't need to use the density of the final solution.)\n\nThe temperature change of the solution is given by the energy released divided by the heat capacity of the system:\n\n(11.2*10^3 J)/(1322 J/degC) = 8.47 deg C\n\nIf the initial solutions had a temperature of 25 C, the final solution has a temperature of 25 + 8.47 C = 33.47 C.
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
801001
Ok, this is a weird question, but here's my best answer.  The 2 liquids stay separate because of density.  The green liquid is cooler and contains salt, both of which increase its density.  The red liquid is warmer (its molecules are moving faster and are farther apart) so it's less dense and should float on the denser green solution (unless you shake them, which is why the card method is used).
226408
the question as written has no answer.  hydronium is an ion that only exists in a solution, never by itself. so it can't have a melting point or freezing point or a density
813249
Let's take the HCl part first. Since HCl is an acid a solution of HCl contains lots of H+ ions. So when you add HCl the concentration of H+ ions increases making it more probable for H+ ions to colide and react with In- ions than before. To be more precise the reaction In- + H+ -> HIn will become more probable than before, so it will dominate over the HIn -> In- + H+ reaction until a new equilibrium is reached (equilibrium is characterized by the left to right and right to left reactions being equally probable). So in equilibrium the concentration of HIn will be higher after you have addeed HCl.\n\nIt's a very general equilibrium principle called leChateliers principle which comes into play here. The principle can be casted as follows: When you disturb an equilibrium (for instance by adding or removing one of the constituting components) the equilibrium will be shifted in a direction which tries to cancel the change. In this example: We add H+ and the equilibrium tries to remove some of the H+ ions again by shifting the equilibrium to the left (meaning that more H+ reacts with In- to get HIn).\n\nIf you add OH- instead the OH- will neutralize the H+ in the solution (acid + base gives water):\nH+ + OH- -> H2O\nso you are effectively removing H+ and according to leChateliers principle the equilibrium will try to create new H+ ions by shifting to the right (so you will have less HIn and more In- than before you added OH-). The microscopic explanation is again that removing H+ makes the reaction In- + H+ -> HIn less probable than before meaning that the reaction HIn -> In- + H+ will be dominating until a new equilibrium is reached so some HIn will effectively be turned into In-.\nHope this helps.
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.
917250
I'm going off memory here so be sure to check this.  I think the formula is the same, but the structure may be different.  Just trying to point you in the right direction.
814756
or_try_this is correct.  It is dense, and a 30-inch column of mercury fits in a laboratory better than a 34-foot column of water.  It is stable and doesn't evaporate at room temperature.  Its thermal expansion can be compensated for, but it's usually small compared to the pressure differences it's measuring.  It looks scientific, too.
775471
I think the cooling effect is the same as heating effect, but opposite of it. If so, then find out the coefficient per degree Celcius of that particular material.\n\nThe formula will be = difference in volume change / the original volume x difference in temperature change = coefficient per degree Celcius. Just reduce the minus (-) sign and treat it as positive even though we know it's reduced.\n\nThen find out if this material is one straight line/rod. If it's in square, then the volume change will be 2x. If cubic, 3x.\n\nDifferrent material have different coefficient. Liquids, solids, gases... Even different solids have different coefficient.\n\nAlso pay attention to...\nDensity = Mass/Volume\n\nI can't guarantee the accuracy but I hope this helps.
58581
Not sure what you mean by "best specific gravity".  I will assume you mean the "best estimate" of the specific gravity.  Also, a newer form of the word "specific gravity" is "relative density".  Both refer to the density of one substance with respect to another substance (where the other substance is often pure water).\n\nOK, so battery acid is often concentrated sulfuric acid (for lead acid batteries).  The relative density of this acid depends on several things such as: 1) the actual acid (could also be nitric acid, hydrochloric acid, etc.), 2) the concentration of the acid, 3) the temperature of the acid solution, 4) the atmospheric pressure, and 5) the humidity.\n\nSo, the short answer is, you have not provided enough information to give a complete answer, and yes the temperature matters.
190244
Specific Heat Capacity\nThe specific heat capacity of a solid or liquid is defined as the heat required to raise unit mass of substance by one degree of temperature. This can be stated by the following equation: \n\nQ= mc(changes of T)\nwhere, \nQ= Heat supplied to substance, \nm= Mass of the substance, \nc= Specific heat capacity, \nT= Temperature rise.
538049
Yes.  In fact protons even from two different elements are indistinguishable.  This is a fundamental property that is quite interesting, and it results in some very odd behavior.
827524
Easier compared to one with a higher heat capacity:\n\nH/(m*Cp) = (delta T), where m is the mass of the object, H is the heat that must be added (or removed) to raise (or lower) its temperature, delta T is its change in temperature, and Cp is its heat capacity. From the above formula, you can see that larger values of Cp result in small temp. changes, for the same heat amount.
1122699
Depends on what the molarity of the concentrated acid is and what the final concentration you want.\n\ntypical formula is: (final volume x final concentration) / initial concentration = initial volume
66611
No the density would be the same. The sample size does not affect the density. The Volume and Mass grow at proportional rates, so it doesn't affect density.