Author Topic: ANY DOUBTS HERE!!!  (Read 34503 times)

Offline Saladin

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ANY DOUBTS HERE!!!
« on: April 04, 2010, 04:28:00 pm »
Hey u guys, post all ur doubts here, and I will do my best to answer them form u.

Just trying out the tex editor.

cos(x)=\pm\sqrt{\frac{3}{4}}

Offline Sue T

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Re: ANY DOUBTS HERE!!!
« Reply #1 on: April 06, 2010, 04:55:15 pm »
i hope chemistry doubts go here as well
past paper says:

 The equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol and ethanoic
acid, C2H5OH + CH3CO2H CH3CO2C2H5 + H2O, at 60(degree)C is 4.00.
When 1.00 mol each of ethanol and ethanoic acid are allowed to reach equilibrium at 60(degree)C, what
is the number of moles of ethyl ethanoate formed?
A 1/3
B 2/3
C 1/4
D 3/4
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Offline 3ishakay

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Re: ANY DOUBTS HERE!!!
« Reply #2 on: April 06, 2010, 05:55:25 pm »
i hope chemistry doubts go here as well
past paper says:

 The equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol and ethanoic
acid, C2H5OH + CH3CO2H CH3CO2C2H5 + H2O, at 60(degree)C is 4.00.
When 1.00 mol each of ethanol and ethanoic acid are allowed to reach equilibrium at 60(degree)C, what
is the number of moles of ethyl ethanoate formed?
A 1/3
B 2/3
C 1/4
D 3/4



hey ... dyu noe the anser to tht?
wht paper ist frm?

im thinkn the anser is B = 2/3

this is how i did it ..(any1 tell me if i got it wrong .. plz!!)


                      ethanol   +     ethanoic acid    ----->   ethylethanoate   +     water
                      C2H5OH   +    CH3CO2H         ------->  CH3CO2C2H5     +     H2O

b4 eq.                  1                   1                                   0                      0

aftr eq.                1-x                1-x                                 x                       x




thn Kc= [ethylethanoate][water]
            -----------------------
            [ethanol][ethanoic acid]


so its like  


          [ x]^2           =  4
        --------
          [1-x]^2


thn u ju solv it like in maths ..

and thn u get 2/3 =D

rechek the smae crap..


go


           ( 2/3)^2              =    4!!   TADA!!! :D:D:D
          ----------
            (1/3)^2




« Last Edit: April 06, 2010, 06:01:07 pm by 3ishakay »

nid404

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Re: ANY DOUBTS HERE!!!
« Reply #3 on: April 06, 2010, 06:05:15 pm »
hey Thanks :)

Offline 3ishakay

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Re: ANY DOUBTS HERE!!!
« Reply #4 on: April 06, 2010, 06:10:19 pm »

Offline ruby92

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Re: ANY DOUBTS HERE!!!
« Reply #5 on: April 06, 2010, 06:11:54 pm »
the answer is b??? ???

nid404

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Re: ANY DOUBTS HERE!!!
« Reply #6 on: April 06, 2010, 06:12:49 pm »
the answer is b??? ???

U have doubts??
Wait, I'll check her post agn

Offline ruby92

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Re: ANY DOUBTS HERE!!!
« Reply #7 on: April 06, 2010, 06:17:10 pm »
lool no...i get it..i was checking i didnt read 3ishakay post  :P

Offline 3ishakay

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Re: ANY DOUBTS HERE!!!
« Reply #8 on: April 06, 2010, 06:19:30 pm »
lool no...i get it..i was checking i didnt read 3ishakay post  :P


lol cool ..

By the way im not sure abt my post .
i always use the revers method cus its fastr.

im not evn sure if the math bit turns out ryt .. :P

nid404

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Re: ANY DOUBTS HERE!!!
« Reply #9 on: April 06, 2010, 06:19:41 pm »
Yup she's right :)

Offline halosh92

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Re: ANY DOUBTS HERE!!!
« Reply #10 on: April 06, 2010, 07:15:39 pm »
PHYSICS AS
could u just explain how to solve this question step by step please?

Jill is swimming across a river which has  constant current of "k m/s". if jill heads directly across the river with speed "v m/s"
find her resultant speed and the angle her path makes with a line directly across the river when
k= 2
v=1
 thx alot ;D
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline Saladin

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Re: ANY DOUBTS HERE!!!
« Reply #11 on: April 06, 2010, 07:18:03 pm »
really sorry u guys, my electricity was out, so i cud not ansa.

Offline Saladin

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Re: ANY DOUBTS HERE!!!
« Reply #12 on: April 06, 2010, 07:23:09 pm »
PHYSICS AS
could u just explain how to solve this question step by step please?

Jill is swimming across a river which has  constant current of "k m/s". if jill heads directly across the river with speed "v m/s"
find her resultant speed and the angle her path makes with a line directly across the river when
k= 2
v=1
 thx alot ;D


this is a simple trigonometric question.

electricity going again

srry cud not reply

cant scan the thing

will wen i can

thanks


hope u guys dnt mind.

« Last Edit: April 06, 2010, 07:28:37 pm by The Ultimate Dude 321 »

Offline Sue T

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Re: ANY DOUBTS HERE!!!
« Reply #13 on: April 07, 2010, 09:04:59 am »
4got 2 mention th answer was b
thanx!
Knowledge is knowing a tomato is a fruit; Wisdom is not putting it in fruit salad.

nid404

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Re: ANY DOUBTS HERE!!!
« Reply #14 on: April 07, 2010, 09:51:32 am »
PHYSICS AS
could u just explain how to solve this question step by step please?

Jill is swimming across a river which has  constant current of "k m/s". if jill heads directly across the river with speed "v m/s"
find her resultant speed and the angle her path makes with a line directly across the river when
k= 2
v=1
 thx alot ;D


R= root of k2 + v2
= root of 5

tan theta= k/v=2
theta= 63.43o clockwise  down the stream