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I'd say no because temperature can't be constant after a sharp rise. Secondly, it keeps changing the concentration which causes rise in temperature.Guys, will this graph be correct for this question?
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I don't think this is correct since the reaction is endothermic (the positive enthalpy change says it all). So this means that it will 'absorb' heat energy from the water and it's temperature should fall and the greater is the concentration of the reactant, the more energy all of it will need to react so the more the temperature will drop.Guys, will this graph be correct for this question?
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I don't think so. Cuz that's what the first diagram shows, temp change increases with increase in conc. whether it be endo or exoI don't think this is correct since the reaction is endothermic (the positive enthalpy change says it all). So this means that it will 'absorb' heat energy from the water and it's temperature should fall and the greater is the concentration of the reactant, the more energy all of it will need to react so the more the temperature will drop.
In case of exothermic reactions, YES the graph will be this way.
But that is not the case for endothermic reactions. since there is no rapid heat loss in this case.
That is only in compounds where one element has an oxidation state of 2+ while the other has 1- or vice versa.Guys, I never understood this point. Why do we always multiply the term inside the bracket by 2 when we square the term. this is an example 4(ii):
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That's temperature CHANGE not the temperature itself.I don't think so. Cuz that's what the first diagram shows, temp change increases with increase in conc. whether it be endo or exo
First calculate no. of moles of NaOH, then no. of moles of the acid.Can anybody help me with this? I have no idea how to approach this..
Could you elaborate? I see no difference in the way both questions are approachedPart c) for this question.
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Part e) for the question below
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Why are they done differently?
Sources F/M/17/52
M/J/17/51
Well you see in the 2nd question, the temperature is found by extending another line across the first three points after the sample was addedCould you elaborate? I see no difference in the way both questions are approached
Well you see in the 2nd question, the temperature is found by extending another line across the first three points after the sample was added
While in the first one, the already extrapolated line is used rather than extending another line through the first and second point after the sample was added
The question is that in the first question why is the line with the negative gradient (one showing falling temperature) considered to find the temperature rise unlike in the 2nd question in which the line with the positive gradient (one showing rising temperature) used?
well idk my teacher checked it and he didn't award me a mark for it :/The positive grad aint being used in the second question, the negative grad is, just like in first question. That's how you're supposed to do it. you can never draw a line of best fit with just a few points, only possible with more than 5 pts on graph, so the line with negative grad. in both the cases
View the quoted post below.Problem 1: A saturated aqueous solution of magnesium methanoate, Mg(HCOO)2, has a solubility of approximately 150 g dm-3 at room temperature. Its exact solubility can be determined by titrating magnesium methanote against aqueous potassium manganate (VII).
During the titration, the methanoate ion, HCOO-, is oxidizes to carbon dioxide while the manganate (VII) ion, MnO4-, is reduced to Mn2+.
You are supplied with
A saturated aqueous solution of Mg(HCOO)2
Aqueous potassium manganate (VII), KMnO4, of concentration 0.0200 mol dm-3
(a) (i). write the half equations for the oxidation of HCOO-(aq) to CO2(g) and the reduction of MnO4-(aq) to Mn2+ (aq) in acid solution.
(ii) Using the approximately solubility above, calculate the concentration, in mol dm-3, of the saturated aqueous magnesium methanoate and the concentration of the methanoate ions present in this solution
(Ai: H, 1.0, C, 12.0, O, 16.0, Mg, 24.3)
(iii) In order to obtain a reliable titre value, the saturated solution of magnesium methanoate needs to be diluted.
Describe how you would accurately measure a 5.0cm3 sample of saturated magnesium methanate solution and use it to prepare a solution fifty times more dilute than the saturated solution.
(vi) 1 mol of acidified MnO4- ions reacts with 2.5 mol of HCOO- ions.
25.0 cm3 of the diluted solution prepared in (iii) required 25.5 cm3 of 0.0200 mol dm-3 potassium manganate (VII) solution to reach the end point.
Use this information to calculate the concentration, in mol dm-3, of HCOO- ions in the diluted solution.
(Vii) use your answer to calculate the concentration of the saturated solution of magnesium methanoate
Can someone help me with vii
Because they are asking for concentration of SATURATED solution
And the concentration calculated in previous part is of dilute solution
and saturated solution is 50 times saturated so 50 times r the moles of Mg(HCOO-)2
but u see there r 2 moles of HCOO- in 1 mole of Mg(HCOO-)2
so just multiply it by 50/2 = 25
send me a pic of your graphwell idk my teacher checked it and he didn't award me a mark for it :/
and theres nothing in the marking scheme which helps :/
u sure its supposed to be done that way?
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