Chemistry: Post your doubts here!

[TERMINATOR]

Wa Alaykum Asalam Warahmatullahi Wabarakatuhu!
when i tried the question out i got B: 0.05 moldm-3 which i think you got too...but see here we have to read the question very carefully (this was a question already asked by someone and answered by another someone so that's how i understood)
it says "in the ratio of 15 g : 30 g : 15 g" NOT "in the ratio of 15 : 30 :15" Note: There IS a difference!
therefore:
Ans : A
- the percentage of N is the fertiliser is 15/100 = 0.15%
- in 14g of fertiliser, the amount of N is 14 x 0.15 = 2.1g
- the number of moles of N is 2.1g of N is 2.1/14 = 0.15 mole
- concentration = mole/volume, hence, 0.15/5 = 0.03moldm-3
(Jazak' Allahu Khairan to that person!)

Thank you, jazaakAllahu khairan. May Allah highly reward you for your help and for making others read your amazing quotes mashaAllah.

 

[TERMINATOR]

Assalaamualikum. Any help with this mcq problem will be highly appreciated.

 

[darknessinme]

Assalaamualikum. Any help with this mcq problem will be highly appreciated.

For such questions always refer to the basic shapes of molecules you should already know eg. CH4,BF3,H2O,CO2,NH3

bp/lp=bond/lone pair
Trimethylamine has 3 bp and one lp around the N atom. Comparing it to NH3 molecule, it takes a trigonal pyrimidal shape. BF3 is an electron deficient compound and has 3 bp around B so takes a trigonal planar shape to minimise repulsion.
When they react, a dative covalent bond forms between N and B. There are four bp around both the N and B atoms. Comparing this to CH4 molecule shape, the electron pairs occupy a tetrahedral shape around N and B atoms.
Thus overall the molecule has a similar shape to ethane.
Answer:C

 

[TERMINATOR]

For such questions always refer to the basic shapes of molecules you should already know eg. CH4,BF3,H2O,CO2,NH3

bp/lp=bond/lone pair
Trimethylamine has 3 bp and one lp around the N atom. Comparing it to NH3 molecule, it takes a trigonal pyrimidal shape. BF3 is an electron deficient compound and has 3 bp around B so takes a trigonal planar shape to minimise repulsion.
When they react, a dative covalent bond forms between N and B. There are four bp around both the N and B atoms. Comparing this to CH4 molecule shape, the electron pairs occupy a tetrahedral shape around N and B atoms.
Thus overall the molecule has a similar shape to ethane.
Answer:C

Thank you. That totally helped. May Allah highly reward you.

 

[TERMINATOR]

Asalam-o-Alaikum!
could someone answer this question for me and tell me HOW to do it..please, asap??
When 1 mol of N2O4 gas is allowed to come to equilibrium with NO2 gas under standard conditions, only 20% of the N2O4 is converted to NO2.
N2O4 <> 2NO2 Hr = + 58 kJmol-1
Calculate the value of Kc for this reaction. Assume that the volume of the reaction mixture is 1 dm3.

I think it should be sth like this... in the equilibrium N2O4 is going to be 80%. So, I calculated its mole by 80/92 [92 is the relative molecular mass of N2O4] which is around 0.87. So, from this we can easily calculate the mole of NO2...doing 0.87*2 since NO2 has 2 in front of it in equilibrium, and u get 0.87*2=1.74 Mol. Since the volume is given 1 dm3...that means the concentration is going to be the mole itself.....0.87 mol/dm3 and 1.74 mol/dm3. so now u can find the Kc >>>> [1.74]^2 divided by [0.87]... n u get 3.48 mol/dm3.

Hope this is correct.

 

[Warrior66]

I think it should be sth like this... in the equilibrium N2O4 is going to be 80%. So, I calculated its mole by 80/92 [92 is the relative molecular mass of N2O4] which is around 0.87. So, from this we can easily calculate the mole of NO2...doing 0.87*2 since NO2 has 2 in front of it in equilibrium, and u get 0.87*2=1.74 Mol. Since the volume is given 1 dm3...that means the concentration is going to be the mole itself.....0.87 mol/dm3 and 1.74 mol/dm3. so now u can find the Kc >>>> [1.74]^2 divided by [0.87]... n u get 3.48 mol/dm3.
Hope this is correct.

it obviously makes sense when i read your explanation, but unfortunately it's not the correct answer or the method used.
the method is as the following: initial concentration for N2O4 : NO2 is 1:0 but after the reaction reaches to equilibrium the concentration is 0.8:0.2 so the Kc is [0.4]^2/[0.8] Now the thing i don't understand is where did 0.4 come from...why don't we use o.2 ?? so after putting these numbers into the calculator, the value of Kc would be 0.2 moldm^3 and i don't know how... :/

 

[darknessinme]

Asalam-o-Alaikum!
could someone answer this question for me and tell me HOW to do it..please, asap??
When 1 mol of N2O4 gas is allowed to come to equilibrium with NO2 gas under standard conditions, only 20% of the N2O4 is converted to NO2.
N2O4 <> 2NO2 Hr = + 58 kJmol-1
Calculate the value of Kc for this reaction. Assume that the volume of the reaction mixture is 1 dm3.

1mol N2O4 initially. It says at equilibrium,20% of it has reacted. So 0.2 mol of it has reacted with 0.8mol remaining. Look at the balanced chemical equation. For every 1mol N2O4 that reacts, 2mol of NO2 forms. Therefore if 0.2mol of N2O4 reacts, you clearly produce 0.4mol of NO2. Then you figure out the concentrations(which is dividing by 1 in this case). Finally you work out Kc=0.4^2/0.8

 

[xavier13]

Greetings everyone! Im a newbie in this site. Anyway, anyone has any links for chemistry topical exercises on internet? Truly appreciate it if anyone of u can share with me. Thanks so much!!!

 

[TERMINATOR]

Salaam. Any help with this mcq. .. it will be highly appreciated.

 

[darknessinme]

Salaam. Any help with this mcq. .. it will be highly appreciated.

With cold KMnO4: Alkenes are oxdised to di-ol. So two -OH groups at the C=C bond in cholesterol give three -OH total.
With hot KMnO4: Alkenes split/break at the C=C bond and are oxidised to give products depending on what's attached. But here all you need to know is that the hexagon ring breaks, leaving 2 hexagons.

 

[TERMINATOR]

With cold KMnO4: Alkenes are oxdised to di-ol. So two -OH groups at the C=C bond in cholesterol give three -OH total.
With hot KMnO4: Alkenes split/break at the C=C bond and are oxidised to give products depending on what's attached. But here all you need to know is that the hexagon ring breaks, leaving 2 hexagons.

Thank you very much. May you be highly blessed, ameen.

 

[pakiboy]

Guyz help i want to give phy AS priv. this year with eco and maths through scl but was shocked when I read the decl we make when reg. priv.
I have not submitted a GCE exam entry at any other centre/school except for the above GCE exam entry through the British Council.
read this ........

 

[scouserlfc]

Guyz help i want to give phy AS priv. this year with eco and maths through scl but was shocked when I read the decl we make when reg. priv.
I have not submitted a GCE exam entry at any other centre/school except for the above GCE exam entry through the British Council.
read this ........

Cant see anything and so what happened ???

 

[pakiboy]

in short can a student give some papers privately and other from scl in a single session?

 

[TERMINATOR]

Salaam. Another small mcq question here. Any help will be highly appreciated.

 

[TERMINATOR]

An mcq problem. Plz help

 

[strangerss]

hey can someone help me out please , i need the answers for the end of chapter questions in the chemistry book , but i don't have the CD so how can I get the answers?

 

[M.Kabir Nawaz]

I've read in E.N.Ramsden that H2+I212HI and Kc=[HI]2/[H] But in a cambridge course book I've read this equation(Attachment). Which one is correct and in case both are correct then Kc is equal to its reciprocal. And if it is not then how can be the Kc different for a reversible reaction in same conditions?

 

[M.Kabir Nawaz]

in short can a student give some papers privately and other from scl in a single session?

no there can be only one statement of entry

 

[pakiboy]

no there can be only one statement of entry

many people give some papers privately although this is isnt allowed! i am going to follow the norm