Chemistry: Post your doubts here!

[Xylferion]

View attachment 52146
Answer is C

As the plunger is pushed in, the volume decreases and the pressure increases. Higher pressure favours a fewer number of moles. At equilibrium there's 1 mol PCl5 on the left compared to 2 moles of gases on the right. So as the plunger is pushed in, the position of the equilibrium shifts to the left, producing more PCl5.

If no dissociation were to happen, the reaction would not be in equilibrium. Hence only 2 moles of gas would be present at the end. Since it partly dissociated though, you can take into account the moles of PCl5, since that is present too. So you'd have 3 moles compared to 2 without dissociation. Which agrees with statement 3.

As for statement 1, I believe it's the pressure of the compressed gas that's greater but not what's in the syringe oven. Since the pressure of compressed air is greater than atmospheric pressure, I can't think of any other reason why the statement might be wrong. It has to be the wording.

Hope that helped.

 

[Xylferion]

Why are BOTH increasing?

Thank you for your help!

You already said why, because you still want the reaction to be in equilibrium.

 

[My Name]

You already said why, because you still want the reaction to be in equilibrium.

Yes...okay
So what I said here

Equilibrium means that the rate of forward reaction is equal to the rate of backward reaction.
So they increased the temperature and still want it to be in Equilibrium?
When the temperature is increased the temperature more of products will be formed since forward reaction is exothermic....but since they want equilibrium both sides increase...
Did I get it right ?

was right
Finally...I get it.... Thank you!

 

[Xylferion]

View attachment 52149


Can someone please answer this question for ME PLEASEEEEEEEEEEEEEEE AND EXPLAINN THANKYOU



ANSWER IS B BUT YYYYYYYYYYYYYe

You write out the bromination of propane.
Br2 ---> 2Br・

Br・ + C3H8 ---> C3H7・ + HBr

C3H7・ + Br2 ----> C3H7Br + Br・

Br・ + C3H8 ----> C3H7 ・ + HBr

Then in the termination step, you have two radicals react. So you can have C3H7・ and C3H7・ react to form C6H14.

The catch here is that there's two types of radicals that react. It may look like it's just C3H7・, but the ・ can be attached to 2 different carbons in C3H7. Either the first or the second.

Like so,

CH3CH2CH2・

Or

CH3CH・CH3

Option 1 follows,

CH3CH2CH2・ + CH3CH・CH3

While option 2 follows,

CH3CH・CH3 + CH3CH・CH3

Option 3 cannot be formed because no combination of CH3CH・CH3 and CH3CH2CH2・ would allow for it.

Hope that made sense!

 

[mohammed awadh]

You write out the bromination of propane.
Br2 ---> 2Br・

Br・ + C3H8 ---> C3H7・ + HBr

C3H7・ + Br2 ----> C3H7Br + Br・

Br・ + C3H8 ----> C3H7 ・ + HBr

Then in the termination step, you have two radicals react. So you can have C3H7・ and C3H7・ react to form C6H14.

The catch here is that there's two types of radicals that react. It may look like it's just C3H7・, but the ・ can be attached to 2 different carbons in C3H7. Either the first or the second.

Like so,

CH3CH2CH2・

Or

CH3CH・CH3

Option 1 follows,

CH3CH2CH2・ + CH3CH・CH3

While option 2 follows,

CH3CH・CH3 + CH3CH・CH3

Option 3 cannot be formed because no combination of CH3CH・CH3 and CH3CH2CH2・ would allow for it.

Hope that made sense!

im so weak in this . do u have an easy explaination ? i understood the radical part but applying it was difficult.
thanks really for the tip

 

[Xylferion]

im so weak in this . do u have an easy explaination ? i understood the radical part but applying it was difficult.
thanks really for the tip

With propane, there are 3 carbons. You can have the little dot ・, which is basically an electron, appear on any of the three carbons.

The thing is, if you have it on the first and third carbon, that's the same thing, because you will start numbering the carbons, from the one that has ・.

The only other carbon which you can have the ・ attached to, is the second one.

This gives you only two places to put ・.

During the termination step, you have two combine two radicals. That is two molecules that have ・.

Since ・ can be on the first OR the second carbon, you can form 3 combinations.

First carbon radical ・ + First Carbon radical ・
First carbon radical ・ + Second carbon radical ・
Second carbon radical ・ + Second carbon radical ・

So if you have C3H7・ as your radical. The ・ can be on the first or second carbons.

-C-C・-C-
OR
・C-C-C-

 

[Mahmood Magdy]

Guys I need your help! How do I prepare a solution of sodium nitrate at 25 mmol dm–3 knowing that the molar mass of sodium nitrate is 85 g mol–1.
Thanks in advance

 

[Xylferion]

Guys I need your help! How do I prepare a solution of sodium nitrate at 25 mmol dm–3 knowing that the molar mass of sodium nitrate is 85 g mol–1.
Thanks in advance

Too little information right there. All you've mentioned is the concentration you want and the molar mass you have.

Concentration =[Mass / Molar Mass ] / Volume

You've only mentioned,

25 = [Mass / 85] / Volume

You're missing one more quantity.

 

[mohammed awadh]

With propane, there are 3 carbons. You can have the little dot ・, which is basically an electron, appear on any of the three carbons.

The thing is, if you have it on the first and third carbon, that's the same thing, because you will start numbering the carbons, from the one that has ・.

The only other carbon which you can have the ・ attached to, is the second one.

This gives you only two places to put ・.

During the termination step, you have two combine two radicals. That is two molecules that have ・.

Since ・ can be on the first OR the second carbon, you can form 3 combinations.

First carbon radical ・ + First Carbon radical ・
First carbon radical ・ + Second carbon radical ・
Second carbon radical ・ + Second carbon radical ・

So if you have C3H7・ as your radical. The ・ can be on the first or second carbons.

-C-C・-C-
OR
・C-C-C-

View attachment 52150

View attachment 52151

Thank you so much. so you are basically joining the free radicals right? if so , then does the radical dot give rise to the methyl group branched?

 

[Xylferion]

Thank you so much. so you are basically joining the free radicals right? if so , then does the radical dot give rise to the methyl group branched?

Nope, it's just that when they join, the radical dot continues its chain with the other radical. It all depends on how you connect the radicals together. Much like tetris.

 

[mohammed awadh]

Nope, it's just that when they join, the radical dot continues its chain with the other radical. It all depends on how you connect the radicals together. Much like tetris.

Man you are the best how do people get such brains haha very intrigued

 

[The Chill Master]

M/j/09/21

Q 4 (b) both part can someone please explain???????????

Q (5) (e) part what should be the reasoning???????
Like should we define why we name the compound cis and all other stuff or is it something else?????

Thanks in advance.

 

[qwertypoiu]

M/j/09/21

Q 4 (b) both part can someone please explain???????????

Q (5) (e) part what should be the reasoning???????
Like should we define why we name the compound cis and all other stuff or is it something else?????

Thanks in advance.

I have assumed you managed to find these compounds (ask if you didn't)
Compound C is CH2(OH)CH(OH)CO2H
Compound D is CH3CH2OH
Compound E is CH3CO2H
Note that C contains two alcohol groups (diol), and a carboxylic acid group.
D contains an alcohol group, and E a carboxylic acid group.
i) For C with D:
Since D is an alcohol, it MUST react with the acid group of C to form an ester:
CH2(OH)CH(OH)CO2CH2CH3
ii) For C with E:
Since E is an acid, it MUST react with the alcohol groupS of C to form a DIester:
CH2(O2CCH3)CH(O2CCH3)CO2H
(two E's react with one C)

Note that O2C is the ester group reversed. (CO2)

 

[The Chill Master]

I have assumed you managed to find these compounds (ask if you didn't)
Compound C is CH2(OH)CH(OH)CO2H
Compound D is CH3CH2OH
Compound E is CH3CO2H
Note that C contains two alcohol groups (diol), and a carboxylic acid group.
D contains an alcohol group, and E a carboxylic acid group.
i) For C with D:
Since D is an alcohol, it MUST react with the acid group of C to form an ester:
CH2(OH)CH(OH)CO2CH2CH3
ii) For C with E:
Since E is an acid, it MUST react with the alcohol groupS of C to form a DIester:
CH2(O2CCH3)CH(O2CCH3)CO2H
(two E's react with one C)

Note that O2C is the ester group reversed. (CO2)

Thank's a lot..........

 

[qwertypoiu]

M/j/09/21

Q 4 (b) both part can someone please explain???????????

Q (5) (e) part what should be the reasoning???????
Like should we define why we name the compound cis and all other stuff or is it something else?????

Thanks in advance.

Just saw your second question sorry missed it before.
For cis and trans answers, the explanation should be in terms of the methyl groups or hydrogen atoms being on the same or
opposite sides relative to the C=C bond. So you have to state the reason why you chose to call a compound cis or trans.

 

[Princess Raven]

W09 p42 can some please explain question 1 part b
And question 3c ii only the balancing part and question 8 d
Please

 

[The Chill Master]

Just saw your second question sorry missed it before.
For cis and trans answers, the explanation should be in terms of the methyl groups or hydrogen atoms being on the same or
opposite sides relative to the C=C bond. So you have to state the reason why you chose to call a compound cis or trans.

Oh.
Thanks for the help.

 

[MYLORD]

can anyone help me with this question

 

[Mahmood Magdy]

Too little information right there. All you've mentioned is the concentration you want and the molar mass you have.

Concentration =[Mass / Molar Mass ] / Volume

You've only mentioned,

25 = [Mass / 85] / Volume

You're missing one more quantity.

Oh yeah, sorry for that. But thanks for the equation though.

 

[MYLORD]

anyone ?