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Chemistry: Post your doubts here!

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Can anyone help please and Expla
jwpjeO.png
in this ?
If you look at the 2nd reaction the double bond in aldehyde is being attacked by CH3CH2O-, which is a nucleophile, and it involves formation of a single product, so it is nucleophilic addition
 
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they r not in diff directions.... in beta pleated hydrogen bonds form btw diff polypeptide chains and the r diff polypeptide chains
this diagram is lso given in my TB

You're wrong there. They do run in opposite directions. I checked my notes to verify if I was mistaken.

BetaPleatedSheetProtein.png
 
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please some one help me http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf

Q23 , Q20 , Q15 , Q17 , Q11 , Q5 , Q6 , Q3 ... answer as much as you can .. so that I don't disturb you and inbox me if required thanks ^_^

3 - D
Na3- so gains 3 electrons.
1s2 2s2 2p6

5 B

Hydrogen bonding exists b/w compounds containing N/O/F.

6 A
Almost picked D overlooking ammonia but it'll be A bec ammonia has Hydrogen Bonding.

11 A?
I'd go with A. OH- produced would react with the H+ ions added to form H2O again which would produce even more HOCl. B will maintain a high conc too, but A will maintain a higher concentration.


15 A
Most definitely A.
AlCl3 exists as a dimer at high temperatures, with the Cl lone pair forming a dative bond with Al.

17 C

Group 7 [O] review.

Cl2 + 2I- -> 2Cl- + I2


20 C?

A and D are out. Always hated these messed up nucleophile electrophile terms. I always mix them up and have to associate them with the word 'pedophile' to realize what they mean. :p
I guess it's nucleophilic addition because the aldehyde is undergoing a nucleophilic addition reaction. Idk.

23 C in a heartbeat!

A-> Ethanol CH3CH2OH = 46
B-> Ethene CH2=CH2 = 28
C-> Interesting. It can form CH3CH2BR and up to CBr3CBr3. In any case, it'll be this because Br alone > 64.5. I knew this from the many Br79-81 questions I've done so I selected this right away. Done each step to explain.
D-> Chloroethane CH3CH2Cl = 64.5
 
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3 - D
Na3- so gains 3 electrons.
1s2 2s2 2p6

5 B

Hydrogen bonding exists b/w compounds containing N/O/F.

6 A
Almost picked D overlooking ammonia but it'll be A bec ammonia has Hydrogen Bonding.

11 A?
I'd go with A. OH- produced would react with the H+ ions added to form H2O again which would produce even more HOCl. B will maintain a high conc too, but A will maintain a higher concentration.


15 A
Most definitely A.
AlCl3 exists as a dimer at high temperatures, with the Cl lone pair forming a dative bond with Al.

17 C

Group 7 [O] review.

Cl2 + 2I- -> 2Cl- + I2


20 C?

A and D are out. Always hated these messed up nucleophile electrophile terms. I always mix them up and have to associate them with the word 'pedophile' to realize what they mean. :p
I guess it's nucleophilic addition because the aldehyde is undergoing a nucleophilic addition reaction. Idk.

23 C in a heartbeat!

A-> Ethanol CH3CH2OH = 46
B-> Ethene CH2=CH2 = 28
C-> Interesting. It can form CH3CH2BR and up to CBr3CBr3. In any case, it'll be this because Br alone > 64.5. I knew this from the many Br79-81 questions I've done so I selected this right away. Done each step to explain.
D-> Chloroethane CH3CH2Cl = 64.5
You're awesome <3 That helped me a lot
Thanks tonnes :p <3
 
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kH3DCP.png


Can anyone help me please how did we get this ?!

Oh boy. That's a toughie.

Group II metal nitrate upon heating will produce a metaloxide and NO2 in 1:1

X(NO3)2 -> XO + NO2

Umm. I'd probably tackle it by finding the mols of the metal nitrate and metal oxide in terms of x.

5/(x+(2(14+16*3)=5-3.29/(x+16)
5/x+124 = 1.71/x+16
5x+80 = 1.71x + 212.04
3.29x = 132.04
x = 40.1
so, Calcium.
 
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Oh boy. That's a toughie.

Group II metal nitrate upon heating will produce a metaloxide and NO2 in 1:1

X(NO3)2 -> XO + NO2

Umm. I'd probably tackle it by finding the mols of the metal nitrate and metal oxide in terms of x.

5/(x+(2(14+16*3)=5-3.29/(x+16)
5/x+124 = 1.71/x+16
5x+80 = 1.71x + 212.04
3.29x = 132.04
x = 40.1
so, Calcium.
Thank's Bro :)
 
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Oh boy. That's a toughie.

Group II metal nitrate upon heating will produce a metaloxide and NO2 in 1:1

X(NO3)2 -> XO + NO2

Umm. I'd probably tackle it by finding the mols of the metal nitrate and metal oxide in terms of x.

5/(x+(2(14+16*3)=5-3.29/(x+16)
5/x+124 = 1.71/x+16
5x+80 = 1.71x + 212.04
3.29x = 132.04
x = 40.1
so, Calcium.

The equation for the Group II nitrate thermal decomposotion is as follows: 2X(NO3)2 ----> 2XO + 4NO2 + O2
5.00g 1.71g (all 3.29g)

You could do it using a bit of algebra or through trial and error. So:
if X = Mg, number of moles of Mg(NO3)2 = 5/148.3 = 0.0337 mol.
Therefore mass os MgO produced = 0.0337 x 40.3 = 1.36g, so can´t be Mg.

if X=Ca, number of moles of Ca(NO3)2 = 5/164.1 = 0.0305 mol.
Therefore mass of CaO produced = 0.0305 x 56.1 = 1.71g

Best way will be fastest way! ^^
 
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675
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The equation for the Group II nitrate thermal decomposotion is as follows: 2X(NO3)2 ----> 2XO + 4NO2 + O2
5.00g 1.71g (all 3.29g)

You could do it using a bit of algebra or through trial and error. So:
if X = Mg, number of moles of Mg(NO3)2 = 5/148.3 = 0.0337 mol.
Therefore mass os MgO produced = 0.0337 x 40.3 = 1.36g, so can´t be Mg.

if X=Ca, number of moles of Ca(NO3)2 = 5/164.1 = 0.0305 mol.
Therefore mass of CaO produced = 0.0305 x 56.1 = 1.71g

Best way will be fastest way! ^^

My equation is intentionally unbalanced. All I wanted to show was the X(NO3)2 -> XO in a 1:1

And yes. Trial and error + quick calculator work will be quick too.
 
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