6.03 Calorimetry Lab

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Well, this looks like its using some calculations so what I would do is take my 0.045 M and equal it to the 0.25 mL of NH3 and multiply that by 45.0 mL and multiply it by 10 with an exponent of negative 3. Once all of that is multiplied together we should get an answer of 0.01135 moles of our HCI. Now we can find our “Concentration” Which means we would divide our moles (0.01125) to our vol in liters which is 0.025, once we do that, we get an answer of 0.045M of our NH3. Well, since we are on the topic of pH we know that we can use the formula: pH = -log (H3O+). Then what we would do is plug everything into the formula: pH equals -log (2.4 multiplied by 10 (with an exponent of -5). Once we find the answer to this and we add up all of our calculations, we can come to a conclusion that the answer is: 4.6197 as our pH. …show more content…

First thing you would want to do is try measuring out your volume of your solution, maybe from inside a beaker or flask. You also want to start titrating your solution until its specific endpoint has been gained. You can tell this, by the color that gives off from the solution. This is why we add a base to our concentration so we can eventually give off a color and figure out the titration. Well, I think that our methane gas and hydrogen sulfide gas would start giving off a reaction and the concentration would eventually lower. Since our concentration would begin to lower, so would our reaction rate for our reaction. Although, as our methane gas and hydrogen sulfide gas are being formed with our Hydrogen and Carbon disulfide, this is going to “jump start” a reverse reaction, which means if we start adding more of our substances to the reaction, its going to speed up the process. This will go on and on until we are at an equal rate and eventually in an dynamic

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