Chemistry: Post your doubts here!

[Mstudent]

Guys, will this graph be correct for this question?

 

[Holmes]

Guys, will this graph be correct for this question?
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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.
I guess I made a point.

 

[anastasia grey113]

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.
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.

 

[Mstudent]

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):

 

[Mstudent]

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.
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.

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

 

[anastasia grey113]

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 is only in compounds where one element has an oxidation state of 2+ while the other has 1- or vice versa.
In the above example, there are 2.5x10-3 moles of the salt.
Which also means that there is 1 mole of SO42- ions.
But as you can see in each molecule there r two atoms of Ag1+
so in 1 mole of the salt, there should be 2 moles of Ag1+ right?
So as the concentration of Ag is double the concentration of SO42- and also the salt itself, it has to be multiplied by 2.

 

[anastasia grey113]

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

That's temperature CHANGE not the temperature itself.
Let me give u can example.
Let's suppose u add 1 mole of the salt. The temperature drops to 23 degrees so the change is 25-23 = 2
But if u add two moles, it drops to 21 so the CHANGE should be 25 - 21 = 4
so the change increases but the final temperature decreases.

 

[anastasia grey113]

Part c) for this question.

Part e) for the question below


Why are they done differently?
Sources F/M/17/52
M/J/17/51

 

[ErosKuikel]

Can anybody help me with this? I have no idea how to approach this..

 

[ErosKuikel]

Calculate the ph of buffer formed when 10cm3 of 0.1 mol per dm3 NaOH is added to 10cm3of 0.25 mol per dm3 CH3COOH whose pKa is 4.76

 

[anastasia grey113]

Can anybody help me with this? I have no idea how to approach this..

First calculate no. of moles of NaOH, then no. of moles of the acid.
Out of the total moles of acid, some moles will also react with NaOH to make CH3COONa.
CH3COOH + NaOH ---> CH3COONa + H2O
According to the equation above:
No. of moles of acid reacted to makes salt = No. of moles of NaOH = No. of moles o salt formed (since all reactants and products are in 1:1 ratio)
Then find the no. of moles of acid left by subtracting total no. of moles of salt from no. of moles of acid.
Use the following equation.
pH = pKa + log(Moles of salt/Moles of acid)

 

[ErosKuikel]

Yes .thanks a lot

 

[Mstudent]

Part c) for this question.
View attachment 63353
Part e) for the question below
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Why are they done differently?
Sources F/M/17/52
M/J/17/51

Could you elaborate? I see no difference in the way both questions are approached

 

[anastasia grey113]

Could 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?

 

[Mstudent]

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?

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

 

[anastasia grey113]

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

well 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?

 

[Fareez]

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

 

[anastasia grey113]

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

View the quoted post below.

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

 

[Mstudent]

well 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?

send me a pic of your graph

 

[ErosKuikel]

I have no idea how to approach this question . I dont even know what is happening . Anyone, please care to explain this in detail.....