Miscellaneous - bits questions

11

SANDIPAN CHAKRABORTY ·2010-05-04 09:35:52

@cute_cat: the electron is lost from the nucleus and not from the outer orbits......so its not the case of simple ionisation....

actually the eqn is....

¹n₀ - ⁰e_-1 --------> ¹p₁

so.....

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1

cute_cat ·2010-05-04 09:40:16

In which of the following K_p < K_q?

(a) I₂ (g) [eq] 2I(g)

(b) 2BrCl(g) [eq] Cl₂ (g) + Br₂ (g)

(c) CO(g) + 3H₂(g) [eq] CH₄(g) + H₂O(g)

(d) All of the above

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1

cute_cat ·2010-05-04 09:43:54

Which of the following silver salts is insoluble in water ?

(a) AgClO₄

(b) Ag₂SO₄

(c) AgF

(d) AgNO₃

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1

1.618 ·2010-05-04 09:46:36

Ag₂SO₄?

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1.618 ·2010-05-04 09:49:36

It is insoluble in the question.

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1

cute_cat ·2010-05-04 10:04:28

PhCH₂CHCH₂ + NBS â†’ A

what is the compound A formed?

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1

cute_cat ·2010-05-04 10:05:17

ya its insolubel...and ray is right...but hw do u get the ans???

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1.618 ·2010-05-04 10:08:08

I know it is insoluble. Cant tell you the reason coz havnt stared inorganic yet.

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1

1.618 ·2010-05-04 10:09:35

#69

See, the most stable intermidiate will form. So, Br^- will attack accordingly.

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1

1.618 ·2010-05-04 10:10:22

You may use this as a reference:-

http://www.targetiit.com/iit-jee-forum/posts/major-product-11781.html

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11

SANDIPAN CHAKRABORTY ·2010-05-04 10:41:19

oops...ya sry..did'nt read the qsn properly...sry..deleted the post.....

http://www.csudh.edu/oliver/chemdata/solrules.htm

check this out...... a very good article regarding the solubility of salts....

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1

1.618 ·2010-05-04 10:42:53

Thanx sandi bhaiyya!

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39

Pritish Chakraborty ·2010-05-04 10:44:27

A is Ph-CH(Br)-CH=CH₂

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1.618 ·2010-05-04 10:45:46

Yeah...acc. to most stable intermediate na?

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39

Pritish Chakraborty ·2010-05-04 10:52:38

Yep. Allyl+Benzyl radical which resonates a lot.

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11

SANDIPAN CHAKRABORTY ·2010-05-04 11:12:44

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39

Pritish Chakraborty ·2010-05-04 11:55:24

Q69. The one with the greatest depression in freezing point would have the least FP. Calcium chloride has the highest van't hoff factor, of 3. Hence its depression in freezing point will be maximum, making FP minimum.

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1

ABHI ·2010-05-04 20:47:29

ya...69 (a) for the reason DIPRIT-H just mentioned

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1

ABHI ·2010-05-04 20:49:48

that K_p<K_q...question.....ans (c)

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39

Pritish Chakraborty ·2010-05-05 00:25:16

BITSAT Trigo Refreshers

Q1. Evaluate √3cosec20Â° + sec20Â°

Q2. The general value of @ satisfying both the equations sin@ = -1/2 and tan@ = 1/√3 is?

Q3. The angle of elevation of the top of a TV tower from the points A, B, C in a straight line(horizontal plane) through the foot the tower are @, 2@, 3@ respectively. If AB = a, the height of the tower is?

Q4. In any triangle ABC, if 2cosB = a/c, the triangle is? (right angled, isosceles, etcetc)

Q5. In any triangle ABC, if b + c = 3a, find the value of cot(B/2)cot(C/2).

I got stuck at the second one. Not getting the answer completely. Okay got it, SR helped.

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11

SANDIPAN CHAKRABORTY ·2010-05-05 03:33:19

Q3

is the answer....

h= 2 a tan ² @tan ² @ + 2tan @ - 1

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11

SANDIPAN CHAKRABORTY ·2010-05-05 03:53:44

this is how i proceede...to the best of my class 10 knowledge and a lil from 11(the formula part..)

tan 2@ = hx

==>x = htan 2@

tan @ = ha + x

==> a + x = htan @

or a + htan 2@ = htan @

solving we get

h = 2 a tan ² @tan ² @ + 2 tan @ - 1

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1

ayush_gupta ·2010-05-05 04:03:47

in q no. 55 proceed just as q#33
just take (2âˆš5-7)²ⁿ⁺¹ as f'.
then I+f-f'=2(some integer)
and 0<f<1,0<f'<1,
thus -1<f-f'<1. since f-f' is an integer impies f-f'=0.
thus I=2k
thus integral part of (7+2âˆš5)²ⁿ⁺¹ is even

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SANDIPAN CHAKRABORTY ·2010-05-05 04:29:06

@ayush:

in the 3rd line ...why is it ..."--" f' (minus f')

also 2âˆš5-7 dont lie between 0 & 1

its apprx (-2.5) ....

so....[12][12][12][12][12][12][12][12][12][12]

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1

cute_cat ·2010-05-05 05:02:06

@ all thanks a lot...bt jo queries reh gayi hain...unhe b plz solve kar do...

@ ayush...hw do u get this.....and 0<f<1,0<f'<1,
thus -1<f-f'<1.

@ abhirup, hw do u solve that K_p < K_q???

@ sandipan, thanks for that link

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1

1.618 ·2010-05-05 05:37:27

Sandi bhai correct...
Trigo ones are simple...need a little manipulation and a couple of formulae.

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39

Pritish Chakraborty ·2010-05-05 05:45:30

Lol sandipan... h = asin2@ :P

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11

SANDIPAN CHAKRABORTY ·2010-05-05 09:27:22

and what will happen to the other carbon...NBS added Br at allylic position..

so i would also go with Pritish's answer....

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1

cute_cat ·2010-05-05 09:31:41

and someone explain post 84 plzzzzzz

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39

Pritish Chakraborty ·2010-05-05 10:58:48

sandi, in your figure, we first look at triangle EBA.
Angle AEB = 2@ - @ = @ = Angle EAB. (Exterior angle property)
Hence the triangle is isosceles. This means EB = a as well.
Now in triangle EBD,
sin2@ = h/EB = h/a
=> h = asin2@.

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