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Chemistry: Post your doubts here!

qwertypoiu

qwertypoiu

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Q4:
assume partial pressure of CO is x. Then partial pressure of H2O must also be x. Therefore:
Kp = (x*x)/(10,000*10,000) = 0.288
x^2 = 2880000000
x = 53665.63146 = 5.37kPa = A.
Q5:
The answer is B. Adding 0.1 moldm-3 of H2O2 makes the solution more dilute, and thus the intial reaction rate of Y was slower. However, adding more H2O2 caused more oxygen to be given out overall at the end of the reaction, as the graph shows.
Q9:
C3H7OH + 4.5O2 --> 3CO2 + 4H2O
0.1 mole of acid was burned. 0.45 mole of oxygen must have been used up therefore. Volume = 0.45*24 = 10.8 dm3. 0.3 mole of CO2 was relaesed as well. 0.3*24 = 7.2dm3
12dm3 of oxygen existed initially, 10.8 of it was consumed, so 12-10.8=1.2dm3 left, plus 7.2 of co2 = 8.4dm3. Therefore B.

Sorry gtg too many to answer :)
 
qwertypoiu

qwertypoiu

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Farhad221 said:
Asslamoalaikum
I need help with this question
Click to expand...
Answer is D. Only student Y is correct. All the carbon atoms bonded in the ring have tetrahedral structure, and are sp3 hybridised. this is because every carbon is attached to four groups. So like when you draw optical isomers, in fact, every carbon with 4 substituents have a 3 dimensional tetrahedral structure. Therefore, they cannot be on the same plane. As for cis-trans, this is obvious, whenever a C=C double bond exists, a cis-trans isomerism is possible (not always but usually.)


EDIT: Sorry, answer is B actually. The compound is prevented from having a cis-trans isomer, because of the ring.
 
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princess Anu

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image.jpg is this lab preparation for chlorine correct? But haven't we studied Nacl with H2so4 makes HCL and not Cl2?:/
 
S

shazmina

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qwertypoiu said:
Answer is D. Only student Y is correct. All the carbon atoms bonded in the ring have tetrahedral structure, and are sp3 hybridised. this is because every carbon is attached to four groups. So like when you draw optical isomers, in fact, every carbon with 4 substituents have a 3 dimensional tetrahedral structure. Therefore, they cannot be on the same plane. As for cis-trans, this is obvious, whenever a C=C double bond exists, a cis-trans isomerism is possible (not always but usually.)
Click to expand...
The answer is B for that questions .. sorry maybe u might haven't noticed it ... :/
 
Metanoia

Metanoia

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qwertypoiu said:
Answer is D. Only student Y is correct. All the carbon atoms bonded in the ring have tetrahedral structure, and are sp3 hybridised. this is because every carbon is attached to four groups. So like when you draw optical isomers, in fact, every carbon with 4 substituents have a 3 dimensional tetrahedral structure. Therefore, they cannot be on the same plane. As for cis-trans, this is obvious, whenever a C=C double bond exists, a cis-trans isomerism is possible (not always but usually.)
Click to expand...

The part on the cis-trans is incorrect. We are talking of cycloalkenes, for ring structures less than 8 carbons, we will not expect to see the trans isomer due to the strain on the bonds.

shazmina said:
The answer is B for that questions .. sorry maybe u might haven't noticed it ... :/
Click to expand...
 
Metanoia

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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)

princess Anu said:
View attachment 51412
Click to expand...

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+)
 
Metanoia

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princess Anu said:
can someone explain part e...
Click to expand...

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%
 
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