Chemistry: Post your doubts here!

[Eugene99]

ans is D but I get 0.003 or something...help!

 

[My Name]

View attachment 59614
ans is D but I get 0.003 or something...help!

pV=nRT
so converting them to the specific units we get:
103000 x 5.37x10^-3=mass/16 (which is the Mr( x 8.31 x 333
so then multiply the p,V and Mr and you get 8849.76.
Divide that by the RT which is 2767.23.
And you get 3.198 which is aprox 3.2.

In other words equation for finding mass equation becomes:
mass=pV Mr/RT.

Did you get it? :3

 

[Eugene99]

pV=nRT
so converting them to the specific units we get:
103000 x 5.37x10^-3=mass/16 (which is the Mr( x 8.31 x 333
so then multiply the p,V and Mr and you get 8849.76.
Divide that by the RT which is 2767.23.
And you get 3.198 which is aprox 3.2.

In other words equation for finding mass equation becomes:
mass=pV Mr/RT.

Did you get it? :3

Got it! Actually I didn't convert kPa into Pa, that made my answer wrong
Mr is 15 not 16 which means we get exactly 3.2, no need to round off
thank you

 

[My Name]

Got it! Actually I didn't convert kPa into Pa, that made my answer wrong
Mr is 15 not 16 which means we get exactly 3.2, no need to round off
thank you

Ohkay
It is methane it should be 16.
Welcome.

 

[Zoha Ali]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

Question 19? how can the 3rd one have an optically active isomer?

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

Question 19? how can the 3rd one have an optically active isomer?

For the III one, this is the optical isomer possible, with the C with asterisk being the chiral carbon.

 

[drowning-in-studies]

A solution of Sn2+ ions will reduce an acidified solution of MnO4 – ions to Mn2+ ions. The Sn2+ ions are oxidised to Sn4+ ions in this reaction. How many moles of Mn2+ ions are formed when a solution containing 9.5 g of SnCl 2 (Mr: 190) is added to an excess of acidified KMnO4 solution?
A 0.010 B 0.020 C 0.050 D 0.125

 

[drowning-in-studies]

Lactic acid (2-hydroxypropanoic acid), CH3CH(OH)CO2H, is found in sour milk. Which reaction could occur with lactic acid?
A CH3CH(OH)CO2H + CH3OH → CH3CH(OCH3)CO2H + H2O
B CH3CH(OH)CO2H + HCO2H → CH3CH(O2CH)CO2H + H2O
C CH3CH(OH)CO2H + NaHCO3 → CH3CH(ONa)CO2H + H2O + CO2
D CH3CH(OH)CO2H + Cl 2 → CH3CH(Cl)CO2H + HOCl

 

[drowning-in-studies]

Use of the Data Booklet is relevant to this question.
2.30g of ethanol were mixed with an excess of aqueous acidified potassium dichromate(VI). The reaction mixture was then boiled under reflux for one hour. The desired organic product was then collected by distillation. The yield of product was 60.0%. What mass of product was collected?
A 1.32g B 1.38g C 1.80g D 3.20g

 

[Saad the Paki]

Use of the Data Booklet is relevant to this question.
2.30g of ethanol were mixed with an excess of aqueous acidified potassium dichromate(VI). The reaction mixture was then boiled under reflux for one hour. The desired organic product was then collected by distillation. The yield of product was 60.0%. What mass of product was collected?
A 1.32g B 1.38g C 1.80g D 3.20g

Oxidation of ethanol under reflux will give us ethanoic acid as our product. CH3CH2OH+[O]-->CH3COOH.
Now we know one mole of ethanol gives one mole of ethanoic acid. So moles of ethanol: 2.3/46=0.05 moles. This is also the moles of ethanoic acid. Now we can find the theoretical yield: Mr×moles=0.05×60=3G
The formula for yield is: (practical yield/theoretical yield)
Yield is given in the question,we found the theoretical yield and we need the practical yield
Therefore : 60%×3g=1.8g. Ans is C

 

[Saad the Paki]

A solution of Sn2+ ions will reduce an acidified solution of MnO4 – ions to Mn2+ ions. The Sn2+ ions are oxidised to Sn4+ ions in this reaction. How many moles of Mn2+ ions are formed when a solution containing 9.5 g of SnCl 2 (Mr: 190) is added to an excess of acidified KMnO4 solution?
A 0.010 B 0.020 C 0.050 D 0.125

The only difficult thing here is to form the equations, which u have to do by balancing the oxidation numbers.
Once the equation is there, find the moles of Sn2+, and then do the ratio thing.

 

[Saad the Paki]

Lactic acid (2-hydroxypropanoic acid), CH3CH(OH)CO2H, is found in sour milk. Which reaction could occur with lactic acid?
A CH3CH(OH)CO2H + CH3OH → CH3CH(OCH3)CO2H + H2O
B CH3CH(OH)CO2H + HCO2H → CH3CH(O2CH)CO2H + H2O
C CH3CH(OH)CO2H + NaHCO3 → CH3CH(ONa)CO2H + H2O + CO2
D CH3CH(OH)CO2H + Cl 2 → CH3CH(Cl)CO2H + HOCl

The first reaction is wrong as the alcohol will only react with CO2H group to form ester, it can't react with another alcohol.
Third reaction is wrong again as only carboxylic acids react with carbonates to release CO2, not alcohols.
Fourth is wrong cuz alcohols do not react with elemental chlorine, they react with chlorine compounds like HCL,PCL3 etc.
The ans is therefore B, because the second reaction is an esterification, the OH has been replaced by part of the carboxylic acid.

 

[drowning-in-studies]

Oxidation of ethanol under reflux will give us ethanoic acid as our product. CH3CH2OH+[O]-->CH3COOH.
Now we know one mole of ethanol gives one mole of ethanoic acid. So moles of ethanol: 2.3/46=0.05 moles. This is also the moles of ethanoic acid. Now we can find the theoretical yield: Mr×moles=0.05×60=3G
The formula for yield is: (practical yield/theoretical yield)
Yield is given in the question,we found the theoretical yield and we need the practical yield
Therefore : 60%×3g=1.8g. Ans is C

Thank you That was a nice explanation! Do you know if this is out of the syllabus?

 

[drowning-in-studies]

The first reaction is wrong as the alcohol will only react with CO2H group to form ester, it can't react with another alcohol.
Third reaction is wrong again as only carboxylic acids react with carbonates to release CO2, not alcohols.
Fourth is wrong cuz alcohols do not react with elemental chlorine, they react with chlorine compounds like HCL,PCL3 etc.
The ans is therefore B, because the second reaction is an esterification, the OH has been replaced by part of the carboxylic acid.

Oh...okay thanks

 

[drowning-in-studies]

The only difficult thing here is to form the equations, which u have to do by balancing the oxidation numbers.
Once the equation is there, find the moles of Sn2+, and then do the ratio thing.

Thank you so much!

 

[Saad the Paki]

Thank you That was a nice explanation! Do you know if this is out of the syllabus?

Your welcome, and no we still have this in the AS syllabus
Actually this was in the igcse syllabus so that's probably why it isn't mentioned in the AS syllabus

 

[shehroze iqbal]

In an experiment various masses of the sodium salt of the acid, NaX, are added to separate portions of 100 cm3 of HX with stirring. After each addition the pH of the solution obtained is measured.

The graph shows that a pH of 3.86 is obtained when 1.12 g of NaX is added to 100 cm3 of HX. Remember that pKa of HX is also 3.86. Use this information to calculate the relative molecular mass, Mr , of HX. Show your working. [Ar : H, 1.0; C, 12.0; O, 16.0; Na, 23.0]

 

[qwertypoiu]

How do you do Q 5 and 6?
View attachment 59613

Q6. In any hydrolysis reaction, the catalyst that can be used are of two types: acid and base.

Answer should therefore be C, to increase reaction rate by catalysis.

 

[holoholo]

When NH3(aq) is added to a green solution containing Ni2+(aq) ions, a grey-green precipitate is formed. This precipitate dissolves in an excess of NH3(aq) to give a blueviolet solution. Suggest an explanation for these observations, showing your reasoning and including equations for the reactions you describe.

The green ppt. is Ni(OH)2. How are we supposed to get that ?

 

[qwertypoiu]

In an experiment various masses of the sodium salt of the acid, NaX, are added to separate portions of 100 cm3 of HX with stirring. After each addition the pH of the solution obtained is measured.

The graph shows that a pH of 3.86 is obtained when 1.12 g of NaX is added to 100 cm3 of HX. Remember that pKa of HX is also 3.86. Use this information to calculate the relative molecular mass, Mr , of HX. Show your working. [Ar : H, 1.0; C, 12.0; O, 16.0; Na, 23.0]

You should have mentioned that the concentration of HX is 0.1 mol/dm^3.

Remember the formula for finding pH in a buffer solution?

pH = pKa + log([salt]/[acid])

If the pH is equal to pKa, like in this case, then:

[salt] = [acid]
[NaX] = [HX]

The question paper mentioned [HX] = 0.1mol/dm^3, so:
[NaX] = 0.1mol/dm^3
moles of NaX in 100cm^3 = 0.1 * 100/1000 = 0.01mol
Mr of NaX = 1.12g / 0.01mol = 112g/mol.
So Mr of HX = 112 - 22 = 90g/mol