Chemistry: Post your doubts here!

[hela]

would you please solve

38 A number of alcohols with the formula C4H10O are separately oxidised. Using 70g of the alcohols
a 62% yield of organic product is achieved.
What mass of product could be obtained?
1 42.2g of butanone
2 51.6g of butanoic acid
3 51.6g of 2-methyl propanoic acid

 

[Raiyan3]

Can someone pleasee explain me the the following paper 1 questions
o/n 05 Q1
m/j 05 Q1 & 2
o/n 04 Q3
m/j 04 Q3
o/n 03 Q1
:/

 

[lolpop11]

6 The diagram shows two bottles of liquid oven cleaner.
The oven cleaners contain sodium hydroxide solution. Plan an investigation to show which
oven cleaner contains the highest concentration of sodium hydroxide

 

[littlecloud11]

would you please solve

38 A number of alcohols with the formula C4H10O are separately oxidised. Using 70g of the alcohols
a 62% yield of organic product is achieved.
What mass of product could be obtained?
1 42.2g of butanone
2 51.6g of butanoic acid
3 51.6g of 2-methyl propanoic acid

38) Oxidizing one mole of the alcohol would give one mole of all three products mentioned.
Mr of C4H10O = 74
70 g of alcohol was used so 70/74 = .946 mols were used.

Butanone - C4H8O; Mr = 72
If .946 mols of alcohol was used then .946 mols of butanone should be produced, meaning .946* 72 = 68.1 g of butanone should have been produced theoretically.
So the yield of butanone is= (42.4/ 68.1) *100 = 61.96% approx. 62%

Butanoic acid - C4H8O2; Mr = 88
If .946 mols of alcohol was used then .946 mols of butanoic should be produced, meaning .946* 88 = 83.2 g of butanoic acid should have been produced theoretically.
So yield of butanoic acid = (51.6/83.2) * 100 =61.98% approx. 62%

2-methyl propanoic acid- C4H7O2; Mr =87
If .946 mols of alcohol was used then .946 mols of 2-methyl propanoic acid should be produced, meaning .946* 87 = 82.3 g of 2-methyl propanoic acid should have been produced theoretically.
So yield of 2-methyl propanoic acid = (51.6/82.3)* 100 = 62.7% approx. 62%

Therefore, all three are likely masses of the products considering all of them gives 62% yield.

 

[hela]

would you please solve

 

[hassankhan]

plz help me with these questions:
Q1) The value of pV is plotted against p for two gases , an ideal gas and a non ideal gas where p is the pressure and V is the volume of the gas.
Which of the following gases show the greatest deviation from ideality
A)ammonia
B)ethene
C)methane
D)nitrogen

Q2)The density of ice is 1.00g/cm^3

What is the volume of steam produced when 1cm^3 of ice is heatedto 323 centrigrade (596 K)at a presssure of one atm(101kPa)

 

[hela]

CAN SOME SOLVE THESE

 

[hela]

PLEA SE SOLVE

 

[hela]

SOME SOLVED PAPER 5 PLEASE

 

[sweetsis]

can u please post the marking scheme of june 2002 chem theory paper

 

[hela]

can u please post the marking scheme of june 2002 chem theory paper

HERE IS TH MS JUNE 2002 P2

 

[MONJUR HASAN NISHAD]

I need help. I can't understand these two MCQs. i) Which of the enthalpy changes of the following reactions can only be obtained by Hess's law. 1 The hydration of anhydrous copper sulphate to form crystals of CuSO4.5H2O 2. The formation of methane from its elements 3. The combustion of glucose, C6H12O6 ii) Sodium ions can be formed from sodium atoms. Na(s) ---> Na+(g) Which quantities are required to calculate the enthalpy change of formation of gaseous sodium ions? 1 enthalpy change of atomisation of sodium 2 first ionisation energy of sodium 3 enthalpy change of formation of sodium.

 

[MONJUR HASAN NISHAD]

Can someone also please clarify May/June 2006 paper1 question number 38? Thanks in advance.

 

[daredevil]

CAN SOME View attachment 22063 SOLVE THESE

AlCl3 ... its that concept about the charge density of Al being too high that it attains a covalent character by attracting the electrons more towards itself.... if u don't get it them reply and i'll explain in detail

 

[daredevil]

PLEAView attachment 22064 SE SOLVE

what is the answer given in the ms?

 

[daredevil]

Can someone also please clarify May/June 2006 paper1 question number 38? Thanks in advance.

refer to the organic reactions and see the HBr reactions, you'll find one that makes that benzene ring thing in option 1 and of course the halogenoalkane from option 2 .... option 3 however will not be selected because the H in the OH group is not replaced when a reaction happens, this whole group either goes or it doesn't change.... i hope i got it right.

 

[daredevil]

I need help. I can't understand these two MCQs. i) Which of the enthalpy changes of the following reactions can only be obtained by Hess's law. 1 The hydration of anhydrous copper sulphate to form crystals of CuSO4.5H2O 2. The formation of methane from its elements 3. The combustion of glucose, C6H12O6 ii) Sodium ions can be formed from sodium atoms. Na(s) ---> Na+(g) Which quantities are required to calculate the enthalpy change of formation of gaseous sodium ions? 1 enthalpy change of atomisation of sodium 2 first ionisation energy of sodium 3 enthalpy change of formation of sodium.

i odn't get these answers either... tag me to the answers wen u get them?? ^_^

 

[hela]

what is the answer given in the ms?

answer is C in ms

 

[daredevil]

answer is C in ms

yeah thats what i thought.... CN has a triple bond.. that means all three of the N's valence electrons are bonded but 1 electron on C is remaining that is the cyanide ion... so 3 PAIRS of electrons are in bonding. other than that one pair in the s shell of each atom is unbonded. got it??

 

[MONJUR HASAN NISHAD]

i odn't get these answers either... tag me to the answers wen u get them?? ^_^

I have no idea about the first question but I figured out the second one. In order for Na(s) to become Na+(g), two separate reactions has to take place. Reaction 1 : Na(s) --- Na(g) [enthalpy change of atomisation of Na] Reaction 2 : Na(g) --- Na+(g) [first ionisation energy of Na]. The enthalpy change of the reaction can hence be calculated by adding enthalpy change of atomisation of Na and the first ionisation energy of Na. The answer is therefore B (Option 1 and 2). As for option 3, enthalpy change of formation of any element is 0. So option 3 won't be considered. If anyone can figure out the first question, let me know!