Chemistry: Post your doubts here!

[Metanoia]

hey metonia, I am done with q6 but i was still wondering for question 5, why can't option B and D be the answers?
thanks in advance

What are your carbon products for B and D? And their bond angles?

View attachment 51215

why C? ._.

What wrong answer did you choose?

 

[cashew]

What are your carbon products for B and D? And their bond angles?


What wrong answer did you choose?

B and then i cut it and chose D :/

 

[Metanoia]

View attachment 51215

why C? ._.

B and then i cut it and chose D :/

Covalent forces is the bond of an I atom to another I atom.
It is very strong and isn't broken during such heating, i.e. the I2 molecule is not broken into two iodine atoms.

Induced dipole-dipole is the weak attraction of an I2 molecule to another I2 molecule. Heating overcomes these forces and separate an I2 molecule from another molecule.

I couldn't find illustrations of iodine molecules, so will use CO2 as example.

 

[cashew]

Covalent forces is the bond of an I atom to another I atom.
It is very strong and isn't broken during such heating, i.e. the I2 molecule is not broken into two iodine atoms.

Induced dipole-dipole is the weak attraction of an I2 molecule to another I2 molecule. Heating overcomes these forces and separate an I2 molecule from another molecule.

I couldn't find illustrations of iodine molecules, so will use CO2 as example.

Woah, thanks a lot bro

 

[ashcull14]

In the gas phase, aluminium and a transition element require the same amount of energy to form one mole of an ion with a 2+ charge.
What is the transition element?

 

[cashew]

11 In some fireworks there is a reaction between powdered aluminium and powdered barium nitrate in which heat is evolved and an unreactive gas is produced. What is the equation for this reaction?
A 2Al + Ba(NO3)2 → Al2O3 + BaO + 2NO
B 4Al + 4Ba(NO3)2 → 2Al2O3 + 4Ba(NO2)2 + O2
C 10Al + 3Ba(NO3)2 → 5Al2O3 + 3BaO + 3N2
D 10Al + 18Ba(NO3)2 → 10Al(NO3)3 + 18BaO + 3N2
why doesn aluminium form Al2O3 instead of Al(NO3)3?

 

[Metanoia]

In the gas phase, aluminium and a transition element require the same amount of energy to form one mole of an ion with a 2+ charge.
What is the transition element?

I suppose this is an MCQ question? Use the data booklet and find the sum of 1st I.E and 2nd I.E of Al (577+1820 = 2397) , then see which of the transition metals give the same result.

11 In some fireworks there is a reaction between powdered aluminium and powdered barium nitrate in which heat is evolved and an unreactive gas is produced. What is the equation for this reaction?
A 2Al + Ba(NO3)2 → Al2O3 + BaO + 2NO
B 4Al + 4Ba(NO3)2 → 2Al2O3 + 4Ba(NO2)2 + O2
C 10Al + 3Ba(NO3)2 → 5Al2O3 + 3BaO + 3N2
D 10Al + 18Ba(NO3)2 → 10Al(NO3)3 + 18BaO + 3N2
why doesn aluminium form Al2O3 instead of Al(NO3)3?

Recall that most metal nitrates decompose under heat to produce metal oxides, so even IF Al(NO3)3 is formed during the reaction, it would decompose to form Al2O3.

 

[cashew]

I suppose this is an MCQ question? Use the data booklet and find the sum of 1st I.E and 2nd I.E of Al (577+1820 = 2397) , then see which of the transition metals give the same result.

View attachment 51229
View attachment 51230



Recall that most metal nitrates decompose under heat to produce metal oxides, so even IF Al(NO3)3 is formed during the reaction, it would decompose to form Al2O3.

Thanks again
May God bless you.

 

[Metanoia]

Thanks again
May God bless you.

No problem, you're welcome.

 

[ashcull14]

I suppose this is an MCQ question? Use the data booklet and find the sum of 1st I.E and 2nd I.E of Al (577+1820 = 2397) , then see which of the transition metals give the same result.

View attachment 51229
View attachment 51230



Recall that most metal nitrates decompose under heat to produce metal oxides, so even IF Al(NO3)3 is formed during the reaction, it would decompose to form Al2O3.

THNKS CN U ANS DIS PLEASE?
Q-In the gas phase, aluminium and a transition element require the same amount of energy to form one mole of an ion with a 2+ charge.
What is the transition element?

 

[ashcull14]

 

[cashew]

View attachment 51231

1 and 3?

 

[Metanoia]

THNKS CN U ANS DIS PLEASE?
Q-In the gas phase, aluminium and a transition element require the same amount of energy to form one mole of an ion with a 2+ charge.
What is the transition element?

Hmm..I already answered it in the post, it's cobalt.

 

[Metanoia]

How to do these ??

The 17 cm3 of NaOH used, is that a figure obtained by actually carrying out the experiment or just a number randomly chosen?

If randomly chosen, it might mean that the final answer of the formula of RCO2H would not make sense.

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s13_qp_42.pdf

can someone please help me in question 2 part b.
thank you.

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_42.pdf

can someone please help me in question 2 part b.
thank you.

Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = x mol
moles of CH3COONa = (0.1 - x) mol

[CH3COONa]/[CH3COOH] = x/(0.1 -x)

pH = pKa + lg ([conjugate base]/[acid])
5.5 = - lg (0.0000179) + lg [x/(0.1 -x)]
5.5 = 4.747 + lg [x/(0.1 -x)]
0.75285 = lg [x/(0.1 -x)]

5.6604 = x/(0.1 - x)
0.56604 - 5.6604 x = x
6.6604 x = 0.56604
x = 0.085

 

[DeViL gURl B)]

Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = x mol
moles of CH3COONa = (0.1 - x) mol

[CH3COONa]/[CH3COOH] = x/(0.1 -x)

pH = pKa + lg ([conjugate base]/[acid])
5.5 = - lg (0.0000179) + lg [x/(0.1 -x)]
5.5 = 4.747 + lg [x/(0.1 -x)]
0.75285 = lg [x/(0.1 -x)]

5.6604 = x/(0.1 - x)
0.56604 - 5.6604 x = x
6.6604 x = 0.56604
x = 0.085

Thank you soo much for the explanation. ^.^

 

[DeViL gURl B)]

Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = x mol
moles of CH3COONa = (0.1 - x) mol

[CH3COONa]/[CH3COOH] = x/(0.1 -x)

pH = pKa + lg ([conjugate base]/[acid])
5.5 = - lg (0.0000179) + lg [x/(0.1 -x)]
5.5 = 4.747 + lg [x/(0.1 -x)]
0.75285 = lg [x/(0.1 -x)]

5.6604 = x/(0.1 - x)
0.56604 - 5.6604 x = x
6.6604 x = 0.56604
x = 0.085

but can you please tell me why did you use the volume in the brackets, that is the salt and the acid bracket.
cuz what we've been told is that it should've the concentration. .-.

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

question 1 part c.

please if someone can give a detailed explanation of these sort of questions .. i always get confused .. in the solution or hydration or lattice energy questions.
if someone has made notes on these .. please SHARE.
PLEASE.
Thank you.

 

[Metanoia]

but can you please tell me why did you use the volume in the brackets, that is the salt and the acid bracket.
cuz what we've been told is that it should've the concentration. .-.

In this case , they are numerically same because the concentrations of both solutions are 1 mol/dm3.

I will include extra steps (in green) to the original solution, since its not clear.

Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = vol * concentration = x dm3 * 1 mol/dm3 = x mol
moles of CH3COONa = vol * concentration = (0.1 - x) dm3 * 1 mol/dm3 = (0.1 - x) mol

[CH3COONa] = moles of CH3COONa/volume of buffer = x/0.1
[CH3COOH] = moles of CH3COOH/volume of buffer = (0.1 - x)/0.1

[CH3COONa]/[CH3COOH]
= [x/0.1]/[(0.1-x)/0.1] (cancel the top and bottom 0.1)
= x/(0.1 -x)