Chemistry: Post your doubts here!

[qwertypoiu]

Also in this
When required to prepare 250 cm3 of 0.5 mol/dm3 solution of a crystal of Mr 50g

First realize that we need only 250 cm3, not 1 dm3
Now in 250 cm3 there will be: 50/4 = 12.5 g of the solid

Shouldn't it be 6.25 g ?

Yes correct. Because it's 0.5mol/dm3 not 1mol/dm3.


(PS you should not trust this source)

 

[Abdullahassan_99]

Can someone tell me why the answer is A?

 

[Rizwan Javed]

Can someone tell me why the answer is A?

Use the general gas equation to solve this question.
PV = nRT
Before and after the value is opened, the n (moles of the gas) will remain the same. So make n the subject of the equation:
n = PV/RT
Before opening, let the volume of M be V.
After opening the volume will be 4V. (Volume of M + 3 * Volume of M)

Now simply substitute the values, into the equation above for two situations: before the valve is opened and after the valve is opened. And put them equal:

((10^5)*V)/(8.31*(20+273)) = (P*4V)/(8.31*(100+273))

Solve it to find P (the final pressure)

You'll get the answer A.

 

[nowaisir10]

Thank you so much qwertypoiu

hey are you giving AS this may with variant 12 and 22?

 

[nowaisir10]

Can someone explain question 3iv) for me? I looked for the mark scheme but I couldn't understand it. http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w07_qp_4.pdf[
do you have facebook Id because I want to make a group in facebook who will be giving AS this may with variant 12 and 22

 

[Saad the Paki]

hey are you giving AS this may with variant 12 and 22?

Nope I'm giving 11 and 21

 

[Eugene99]

I can understand until B is added to dilue H2S04 and we get 2hydroxycarboxylic acid but how does adding conc H2SO4, with heat, gets us CH2=CHCO2H and what reaction is this?

 

[qwertypoiu]

View attachment 59849
I can understand until B is added to dilue H2S04 and we get 2hydroxycarboxylic acid but how does adding conc H2SO4, with heat, gets us CH2=CHCO2H and what reaction is this?

This is dehydration of alcohol. It removes an H2O molecule from the alcohol by removing the OH group and an H from the adjacent carbon atom. An alkene is formed. This type of reaction is also known as elimination.
There are other reagents and conditions that can be used like heated alumina (Al2O3) or hot porous pumice (or something like that )

 

[Eugene99]

 

[Eugene99]

This is dehydration of alcohol. It removes an H2O molecule from the alcohol by removing the OH group and an H from the adjacent carbon atom. An alkene is formed. This type of reaction is also known as elimination.
There are other reagents and conditions that can be used like heated alumina (Al2O3) or hot porous pumice (or something like that )

oh thanks a lot! I get that but I was confused because it was a carboxylic acid, wasn't it?

 

[qwertypoiu]

oh thanks a lot! I get that but I was confused because it was a carboxylic acid, wasn't it?

Well yes it did have a carboxylic acid group but I do not know of any reaction that the carboxylic acid group undergoes with H2SO4. However, there WAS the alcohol group so we ignore the acid group and concentrate on the alcohol.

 

[qwertypoiu]

View attachment 59850

The ester group is broken.
OH from water joins the C=O part to form COOH.
H from water joins the O part to form OH.
So when HCO2CH(CH3)R is hydrolysed,
HCO2H and RCH(CH3)OH is formed

 

[ashcull14]

sorry i knw tht im asking the dumbest quebut seriously ive not been through this chapter properly ..can anyone help?? part 2

 

[Eugene99]

Well yes it did have a carboxylic acid group but I do not know of any reaction that the carboxylic acid group undergoes with H2SO4. However, there WAS the alcohol group so we ignore the acid group and concentrate on the alcohol.

Yes exactly!

 

[The Sarcastic Retard]

sorry i knw tht im asking the dumbest quebut seriously ive not been through this chapter properly ..can anyone help?? part 2View attachment 59854

You now know the rate equation.
Rate = [NO][O2] (both have powers of its order) for example NO is 2nd order and O2 is 1st order the rate equation will be [NO]^2.[O2]

You have value of concentrations. Substitute and your life is now easy xD
good luck.

 

[Eugene99]

Sodium hydrogencarbonate is the same as sodium bi-carbonate?

 

[My Name]

Sodium hydrogencarbonate is the same as sodium bi-carbonate?

Yep.

 

[ashcull14]

You now know the rate equation.
Rate = [NO][O2] (both have powers of its order) for example NO is 2nd order and O2 is 1st order the rate equation will be [NO]^2.[O2]

You have value of concentrations. Substitute and your life is now easy xD
good luck.

well i knw tht but i didnt get hw is this the ans in MS
(0.00408 × 27) rate = 0.11 (mol dm–3 s–1) to 2sf

 

[The Sarcastic Retard]

well i knw tht but i didnt get hw is this the ans in MS
(0.00408 × 27) rate = 0.11 (mol dm–3 s–1) to 2sf

Look at this.
I compared 1st and 4th experiment.

In ms its opposite of what i did.
denominator part is numerator and vice versa.

 

[The Sarcastic Retard]

well i knw tht but i didnt get hw is this the ans in MS
(0.00408 × 27) rate = 0.11 (mol dm–3 s–1) to 2sf

In ms its same thing. That 27 came from the comparision of 4th and 1st experiment ratio.