Chemistry: Post your doubts here!

[Mimick]

O

Oka

In the meantime, get on with ur other questions

 

[princess Anu]

The paragraph is trying to bring across two points
1) it is possible for a molecule to be overall non-polar and yet have positive and negative charge regions.
2) there is attraction between the positive region of one molecule and the negative region of another molecule
(however, I think "strong" might be an exaggeration here)



Use the other question where MnO2 is used to react with HCl to form Cl2 and Mn2+

You need a metal oxide that is readily reduced (in this case, from Pb4+ to Pb2+)

so how do these molecules actually differ ( from those having a overall dipole)? I thought they behave exactly like nonpolar molecules and thus they have only induced dipoles and they are insolube in polar solvents.. am I wrong?

 

[princess Anu]

CaMg(CO3)2 + 4HCl --> CaCl2 + MgCl2 + 2H2O + 2CO2

moles of CO2 = 0.45/44 = 0.01022 mol

moles of CaMg(CO3)2 = 0.01022/2

mass of CaMg(CO3)2 = (0.01022/2) x Mr

% purity of CaMg(CO3)2 = (mass of CaMg(CO3)2 /mass of sample) x 100%

Thank you
what should be the ans of part d

 

[The Chill Master]

@awesomaholic101 and anyone
Can you help ? Plz and thnx

 

[Mimick]

Well it seems, someoene already answered...
The Chill Master : you gotta check my reasoning..bcos in question 8 i think the functional group is Ketone and Aldehyde and not alcohol..Bcos according to my understanding if its alcohol the valency of carbon is not filled...draw the displayed formula and see if i make sense

http://postimg.org/image/8jdjxeqi1/

 

[Metanoia]

Thank you
what should be the ans of part d

The stronger the forces of attraction among the ions, the higher the melting points.



The force is directly proportional to the charges and indirectly proportional to the distance between the ions.

In this case, since the compounds are all +2 and -2 charges, the factor that makes a difference is the distance between ions (cation radius + anion radius).

From Mg to Ba, the size of the ion increases due to the increases in electron shells, so distance between ions increases. Attraction forces thus decreases, and so does melting points.

ii) Be more specific, rather than saying that MgO is used in furnaces, write that it is used as a lining for furnaces.

so how do these molecules actually differ ( from those having a overall dipole)? I thought they behave exactly like nonpolar molecules and thus they have only induced dipoles and they are insolube in polar solvents.. am I wrong?

Molecules like CO2
Have polar bonds (permanent dipoles), due to the significant electronegative difference between the atoms
Is still a non-polar molecule (as the dipoles cancel each other out)
Ok solubility in water. Able to form H bonds.
Attract another molecule with permanent dipole-permanent dipole interaction.

Molecules like CH4
Have non-polar bonds (induced dipoles), due to the insignificant electronegative difference between the atoms
Is a non-polar molecule
Poor solubility in water (due to lack of H-bonds)
Attract another molecule with instantaneous dipole-induced dipole interaction.

 

[Metanoia]

@awesomaholic101 and anyone
Can you help ? Plz and thnx

I'll help with Q8 and Q25 as the other questions are already explained.

 

[princess Anu]

The stronger the forces of attraction among the ions, the higher the melting points.

View attachment 51419

The force is directly proportional to the charges and indirectly proportional to the distance between the ions.

In this case, since the compounds are all +2 and -2 charges, the factor that makes a difference is the distance between ions (cation radius + anion radius).

From Mg to Ba, the size of the ion increases due to the increases in electron shells, so distance between ions increases. Attraction forces thus decreases, and so does melting points.

ii) Be more specific, rather than saying that MgO is used in furnaces, write that it is used as a lining for furnaces.



Molecules like CO2
Have polar bonds (permanent dipoles), due to the significant electronegative difference between the atoms
Is still a non-polar molecule (as the dipoles cancel each other out)
Poor solubility in water (due to lack of H-bonds)
Attract another molecule with permanent dipole-permanent dipole interaction.

Molecules like CH4
Have non-polar bonds (induced dipoles), due to the insignificant electronegative difference between the atoms
Is a non-polar molecule
Poor solubility in water (due to lack of H-bonds)
Attract another molecule with instantaneous dipole-induced dipole interaction.

For anything to be soluble in water, we look at the availability of H bonds? or is it just that the molecule has to be polar (according to 'Like dissolves in like' rule)?

 

[Metanoia]

For anything to be soluble in water, we look at the availability of H bonds? or is it just that the molecule has to be polar (according to 'Like dissolves in like' rule)?

limiting the discussion to molecules, it would usually require the molecule to be capable of forming hydrogen bonds with water.
E.g small chain alcohol, esters, ketones, aldehydes, carboxyllic acids

Even CO2, which is a non polar molecule, can dissolve relatively well in water compared to other gases as it can form hydrogen bonds with water. ( I made an error in describing CO2 solubility in previous post).

A molecule that is polar but unable to form Hydrogen bonds with water is usually insoluble.
E.g. CH3Cl

 

[princess Anu]

limiting the discussion to molecules, it would usually require the molecule to be capable of forming hydrogen bonds with water.
E.g small chain alcohol, esters, ketones, aldehydes, carboxyllic acids

Even CO2, which is a non polar molecule, can dissolve relatively well in water compared to other gases as it can form hydrogen bonds with water. ( I made an error in describing CO2 solubility in previous post).

A molecule that is polar but unable to form Hydrogen bonds with water is usually insoluble.
E.g. CH3Cl

how can co2 form hydrogen bonds when it doesn't have an H attached to a electronegative element.. ؟

 

[princess Anu]

can somebody solve thiss

 

[Metanoia]

how can co2 form hydrogen bonds when it doesn't have an H attached to a electronegative element.. ؟

Don't forget that the complete definition also includes a portion where the partially positive H atom is attracted to a lone pair of a neighboring molecule.
The CO2 is responsible for the yellow portion.
http://www.answers.com/Q/Does_carbon_dioxide_form_hydrogen_bonds_with_water

 

[My Name]

Question 8 isnt it Aldehyde it has -CHO :3 just look at it it has Ch3-C=O then -CHO

Question 17 no it isnt its A and I had marked D hehe

Question 25 there is no option 1 AND 3 ....remember? answer is B (I had marked D )
and how CH4+Cl2 is Oxidation reaction

 

[My Name]

I'll help with Q8 and Q25 as the other questions are already explained.

View attachment 51420

Thank you
question 25 answer = B

 

[Metanoia]

Hmm, I'm suprised. Do you have the year of the paper?

 

[My Name]

Well it seems, someoene already answered...
The Chill Master : you gotta check my reasoning..bcos in question 8 i think the functional group is Ketone and Aldehyde and not alcohol..Bcos according to my understanding if its alcohol the valency of carbon is not filled...draw the displayed formula and see if i make sense

http://postimg.org/image/8jdjxeqi1/

I know it HAS to be Aldehyde the answer for that is A!
17 is A Bravo THNX

for question 25 1 AND 3 is not an option remember?

 

[My Name]

Hmm, I'm suprised. Do you have the year of the paper?

And I'm confused.
At the moment I don't but will find it soon.
(You are asking me right?)

 

[Metanoia]

And I'm confused.
At the moment I don't but will find it soon.
(You are asking me right?)

Oh, Cl2 is reduced to HCl. Oxidation state of Cl from 0 to -1

 

[Metanoia]

And I'm confused.
At the moment I don't but will find it soon.
(You are asking me right?)

Oh, Cl2 is reduced to HCl. Oxidation state of Cl from 0 to -1

 

[My Name]

Oh, Cl2 is reduced to HCl. Oxidation state of Cl from 0 to -1

So the other one is oxidised and it becomes redox....so 2 is redox
and so is 1 as you mentioned earlier
the answer is B ( 1 and 2)