English chemistry mnemonics

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English chemistry mnemonics

The elements Hydrogen, Nitrogen, Fluorine, Oxygen, Iodine, Chlorine, and Bromine are always found as diatomic molecules:

HNFOIClBr
HONClFIBr (say HONKLE-fibber)
BrINClHOF (say Brinckle-hoff)
Have No Fear Of Ice Cold Beer
Never Have Fear Of Ice Cold Beer
I Have No Bright Or Clever Friends
I Bring Clay For Our New House
ClIF H Bron
HOFBrINCl Twins (twins because they exist in pairs)
There are seven such elements. The first one is the first element Hydrogen; the rest form a 7 on the periodic table: N, O, F across, then going down Cl, Br, I.
Obsessed Cleveland Brown Fans Have No Intelligence

Oxidation and reduction

Acronyms for oxidation and reduction:

Oxidation Is Losing Electrons, Reduction Is Gaining Electrons: OIL RIG
Electron Loss Means Oxidation: ELMO
Losing Electrons Oxidation, Gaining Electrons Reduction: LEO the lion. GER! or LEO says GER

Oxidation and reduction in the electrodes of an electrochemical cell:

Oxidation takes place at the anode (both start with vowels) reduction takes place at the cathode (both start with consonants).
REDuction at CAThode: RED CAT
ANode for OXidation: AN OX
Anode for Oxidation: Adults Only
Reduction at Cathode: Roman Catholic
At the AN'ode Oxidation Is Gain; and Reduction Is Loss at the CAThoed: AN OIL RIG CAT

The elements that form hydrogen bonds when bonded to hydrogren are Fluorine Oxygen Nitrogen:

Hydrogen is FON! (fun)
I think about hydrogen bonds OFN (often)

Cations are positive, while anions are negative:

The letter t in cation looks like a positive plus sign +: ca+ion
Cations are pawsitive.
An ANION is A Negative ION.
Be POSITIVE about your CATs

Metals form cations in ionic reactions

Metal Cat, Metal Cat, Where is my Metal Cat?

Organic ions:

"A hot carbonium ion is a naked carbonium ion" -- George Olah, in lecture, 1966

The sequence of dicarboxilic acids is Oxalic, Malonic, Succinic, Glutaric, Adipic, Pimelic:

Oh My Stars, Green Apples
Oh My, Such Good Apple Pie

The list of elements from most electronegative to least electronegative begins Fluorine Oxygen Nitrogen Chlorine:

FONCl (phone call)

Free energy equation (Gibbs-Helmholtz equation): ?G = ?H - T?S:

Grades = Hardship - Torture*Suffering, remembered with the terms in alphabetical sequence

Remembering sequence of alcohols,aldehydes,alkenes and alkanes

Many Elderly People Buy Picnic Hampers

Methanol Ethanol Propanol Butanol Pentanol Hexanol

Methanal Ethanal Propanal Butanal Pentanal Hexanal

Methene Ethene Propene Butene Pentene Hexene

Methane Ethane Propane Butane Pentane Hexane

Periodic table

Period 1 of the periodic table contains H and He; period 2 contains Li, Be, B, C, N, O, F and Ne:

HEre LIes BEarded Bill Caught Napping On Friday.
He Likes Beer But Can Not Obtain Food
Harry, He Likes Beer By Cups Not OverFlowing
Harry, He Likes Beer But Cup's Not Over Flowing
Little Betty Boron Chews Nuts On Friday Night
Henry, He Likes Betty But Can Not Offer Flower Necklace
London BBC NO FuN
Liz's Better Babies Could Not Often Feel Nervous

Period 2:

Negress Named Magnolia Alters Sister's Photos Surprisingly Cleverly.
Nelly's Naughty Mg Allows Silly People Some Clatter Around Katie's Carpark
NaMgAl SiPS Chlorine Ardently
Nancy Margaret Always Sing Pop *Song Called "Arren King's Car"

Naughty Maggie Allowed Simon Put Sugar Clothes

Nations Might Ally Since Power Structures Clear the A[i]r [From my own college days, 1968....]

Period 3:

A(r) Kopper Called Scarface Tim, Very Crafty Man, (Fe) i(ro)nCogNito Cunning Zena, Gave Gertie A(s) Severe Bruise

Periods 1, 2 and 3:

H HeLi BeB C NO FNe Na MgAl SiPS ClArK Ca ScTiV CrMn FeCoNiCuZn GaGeAs Se BrKr

Hydrogen Bonds

(
The elements that form hydrogen bonds when bonded to hydrogren are Fluorine Oxygen Nitrogen:

Hydrogen is FON! (fun)
Dicarboxilic Acids (electrodes)
REDuction at CAThode: RED CAT

Oxidation and reduction in the electrodes of an electrochemical cell:
- Oxidation takes place at the anode (both start with vowels), reduction takes place at the cathode (both start with consonants).
- REDuction at CAThode: RED CAT
- ANode for OXidation: AN OX
- Oxidation and Reduction (electrons)
  
  Acronyms for oxidation and reduction:
  - Oxidation Is Losing Electrons, Reduction Is Gaining Electrons: OIL RIG
  - Electron Loss Means Oxidation: ELMO
  - Losing Electrons Oxidation, Gaining Electrons Reduction: LEO the lion. GER! or LEO says GER

Sentences to remember the elements

Periodic Table
Group	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18
Period
1	H																	He
2	Li	Be											B	C	N	O	F	Ne
3	Na	Mg											Al	Si	P	S	Cl	Ar
4	K	Ca	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn	Ga	Ge	As	Se	Br	Kr
5	Rb	Sr	Y	Zr	Nb	Mo	Tc	Ru	Rh	Pd	Ag	Cd	In	Sn	Sb	Te	I	Xe
6	Cs	Ba	*	Hf	Ta	W	Re	Os	Ir	Pt	Au	Hg	Tl	Pb	Bi	Po	At	Rn
7	Fr	Ra	**	Rf	Ha	Sg	Ns	Hs	Mt	Uun	Uuu
/
Lanthanides			*	La	Ce	Pr	Nd	Pm	Sm	Eu	Gd	Tb	Dy	Ho	Er	Tm	Yb	Lu
Actinides			**	Ac	Th	Pa	U	Np	Pu	Am	Cm	Bk	Cf	Es	Fm	Md	No	Lr

bron: Carcamels Homepage

Mnemonics:

1)

First eighteen elements in the periodic table:
Here He Lies Beneath Bed Clothes, Nothing On, Feeling Nervous. Naughty Margaret Always Sighs, "Please Stop Clowning Around."

(H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar.)

2)

First twenty elements in the periodic table:

Hi Hello Little Beryll Brown Cracking Nuts On Friday . Nellie's Naughty Magpie Always Sings Pop Songs Clearly After Killing Cathy

(H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca.)

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Use the formula ALL FROM 9 AND THE LAST FROM 10 to
perform instant subtractions.

For example 1000 - 357 = 643

We simply take each figure in 357 from 9 and the last figure from 10.

So the answer is 1000 - 357 = 643

And thats all there is to it!

This always works for subtractions from numbers consisting of a 1 followed by noughts: 100; 1000; 10,000 etc.

Similarly 10,000 - 1049 = 8951

For 1000 - 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083.
So 1000 - 83 becomes 1000 - 083 = 917

Exercise 1 Tutorial 1

Try some yourself:

1) 1000 - 777 =

2) 1000 - 283 =

3) 1000 - 505 =

4) 10,000 - 2345 =

5) 10,000 - 9876 =

6) 10,000 - 1011 =

7) 100 - 57 =

8) 1000 - 57 =

9) 10,000 - 321 =

10) 10,000 - 38 =

Answers to exercise 1 Tutorial 1 < click

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Index Alphabetical [Index to Pages]

Tutorial 2

Using VERTICALLY AND CROSSWISE you do not need the multiplication tables beyond 5 X 5.

Suppose you need 8 x 7
8 is 2 below 10 and 7 is 3 below 10.
Think of it like this:

The answer is 56.
The diagram below shows how you get it.

You subtract crosswise 8-3 or 7 - 2 to get 5,
the first figure of the answer.
And you multiply vertically: 2 x 3 to get 6,
the last figure of the answer.

That's all you do:

See how far the numbers are below 10, subtract one number's deficiency from the other number, and multiply the deficiencies together.

7 x 6 = 42

Here there is a carry: the 1 in the 12 goes over to make 3 into 4.

Exercise 1 Tutorial 2

Multply These:

1) 8 x 8 =

2) 9 x 7 =

3) 8 x 9 =

4) 7 x 7 =

5) 9 x 9 =

6) 6 x 6 =

Answers to exercise 1 tutorial 2

Here's how to use VERTICALLY AND CROSSWISE for multiplying numbers close to 100.

Suppose you want to multiply 88 by 98.
Not easy,you might think. But with VERTICALLY AND CROSSWISE you can give the answer immediately, using the same method as above

Both 88 and 98 are close to 100.
88 is 12 below 100 and 98 is 2 below 100.

You can imagine the sum set out like this:

As before the 86 comes from subtracting crosswise: 88 - 2 = 86 (or 98 - 12 = 86: you can subtract either way, you will always get the same answer).
And the 24 in the answer is just 12 x 2: you multiply vertically.
So 88 x 98 = 8624

Exercise 2 Tutorial 2

This is so easy it is just mental arithmetic.

Try some:

1) 87 x 98 =

2) 88 x 97 =

3) 77 x 98 =

4) 93 x 96 =

5) 94 x 92 =

6) 64 x 99 =

7) 98 x 97 =

Answers to Exercise 2 Tutorial 2 < click

Multiplying numbers just over 100.

103 x 104 = 10712
The answer is in two parts: 107 and 12,
107 is just 103 + 4 (or 104 + 3),
and 12 is just 3 x 4.

Similarly 107 x 106 = 11342
107 + 6 = 113 and 7 x 6 = 42

Exercise 3 Tutorial 2

Again, just for mental arithmetic

Try a few:

1) 102 x 107 =

2) 106 x 103 =

3) 104 x 104 =

4) 109 x 108 =

5) 101 x123 =

6) 103 x102 =

Answers to exercise 3 Tutorial 2 < click

Top <To top of this page
Index Alphabetical [Index to Pages]

Tutorial 3

The easy way to add and subtract fractions.

Use VERTICALLY AND CROSSWISE to write the answer straight down!

Multiply crosswise and add to get the top of the answer:
2 x 5 = 10 and 1 x 3 = 3. Then 10 + 3 = 13.
The bottom of the fraction is just 3 x 5 = 15.
You multiply the bottom number together.

So:

Subtracting is just as easy: multiply crosswise as before, but the subtract:

Exercise 1 Tutorial 3

Try a few:

Answers to Exercise 1 Tutorial 3 < click

Tutorial 4
A quick way to square numbers that end in 5 using the formula BY ONE MORE THAN THE ONE BEFORE.

75² = 5625

75² means 75 x 75.
The answer is in two parts: 56 and 25.
The last part is always 25.
The first part is the first number, 7, multiplied by the number "one more", which is 8:
so 7 x 8 = 56

Similarly 85² = 7225 because 8 x 9 = 72.

Exercise 1 Tutorial 4

Try these:

1) 45² =

2) 65² =

3) 95² =

4) 35² =

5) 15² =

Answers to Exercise 1 Tutorial 4 < click

Method for multiplying numbers where the first figures are the same and the last figures add up to 10.

32 x 38 = 1216

Both numbers here start with 3 and the last figures (2 and 8) add up to 10.

So we just multiply 3 by 4 (the next number up) to get 12 for the first part of the answer.

And we multiply the last figures: 2 x 8 = 16 to get the last part of the answer.

Diagrammatically:

And 81 x 89 = 7209

We put 09 since we need two figures as in all the other examples.

Exercise 2 Tutorial 4

Practise some:

1) 43 x 47 =

2) 24 x 26 =

3) 62 x 68 =

4) 17 x 13 =

5) 59 x 51 =

6) 77 x 73 =

Answers to Exercise 2 Tutorial 4

Tutorial 5

An elegant way of multiplying numbers using a simple pattern

21 x 23 = 483

This is normally called long multiplication butactually the answer can be written straight downusing the VERTICALLY AND CROSSWISEformula.
We first put, or imagine, 23 below 21:

There are 3 steps:

a) Multiply vertically on the left: 2 x 2 = 4.
This gives the first figure of the answer.
b) Multiply crosswise and add: 2 x 3 + 1 x 2 = 8
This gives the middle figure.
c) Multiply vertically on the right: 1 x 3 = 3
This gives the last figure of the answer.

And thats all there is to it.

Similarly 61 x 31 = 1891

6 x 3 = 18; 6 x 1 + 1 x 3 = 9; 1 x 1 = 1

Exercise 1 Tutorial 5

Try these, just write down the answer:

1) 14 x 21

2) 22 x 31

3) 21 x 31

4) 21 x 22

5) 32 x 21

Answers to Exercise 1 Tutorial 5

Exercise 2a Tutorial 5

Multiply any 2-figure numbers together by mere mental arithmetic!

If you want 21 stamps at 26 pence each you can

easily find the total price in your head.

There were no carries in the method given above.,/p>

However, there only involve one small extra step.

21 x 26 = 546

The method is the same as above
except that we get a 2-figure number, 14, in the
middle step, so the 1 is carried over to the left
(4 becomes 5).

So 21 stamps cost £5.46.

Practise a few:

1) 21 x 47

2) 23 x 43

3) 32 x 53

4) 42 x 32

5) 71 x 72

Answers to Exercise 2a Tutorial 5

Exercise 2b Tutorial 5

33 x 44 = 1452

There may be more than one carry in a sum:

Vertically on the left we get 12.
Crosswise gives us 24, so we carry 2 to the left
and mentally get 144.

Then vertically on the right we get 12 and the 1
here is carried over to the 144 to make 1452.

6) 32 x 56

7) 32 x 54

8) 31 x 72

9) 44 x 53

10) 54 x 64

Answers to Exercise 2b Tutorial 5

Any two numbers, no matter how big, can be

multiplied in one line by this method.

Tutorial 6

Multiplying a number by 11.

To multiply any 2-figure number by 11 we just put
the total of the two figures between the 2 figures.

26 x 11 = 286

Notice that the outer figures in 286 are the 26
being multiplied.

And the middle figure is just 2 and 6 added up.

So 72 x 11 = 792

Exercise 1 Tutorial 6

Multiply by 11:

1) 43 =

2) 81 =

3) 15 =

4) 44 =

5) 11 =

Answers to Exercise 1 Tutorial 6

77 x 11 = 847

This involves a carry figure because 7 + 7 = 14
we get 77 x 11 = 7₁47 = 847.

Exercise 2 Tutorial 6

Multiply by 11:

1) 11 x 88 =

2) 11 x 84 =

3) 11 x 48 =

4) 11 x 73 =

5) 11 x 56 =

Answers to Exercise 2 Tutorial 6

234 x 11 = 2574

We put the 2 and the 4 at the ends.

We add the first pair 2 + 3 = 5.

and we add the last pair: 3 + 4 = 7.

Exercise 3 Tutorial 6

Multiply by 11:

1) 151 =

2) 527 =

3) 333 =

4) 714 =

5) 909 =

Answers to Exercise 3 Tutorial 6

Tutorial 7

Method for dividing by 9.

23 / 9 = 2 remainder 5

The first figure of 23 is 2, and this is the answer.

The remainder is just 2 and 3 added up!

43 / 9 = 4 remainder 7

The first figure 4 is the answer

and 4 + 3 = 7 is the remainder - could it be easier?

Exercise 1a Tutorial 7

Divide by 9:

1) 61 / 9 =         remainder

2) 33 / 9 =         remainder

3) 44 / 9 =        remainder

4) 53 / 9 =        remainder

5) 80 / 9 =        remainder

Answers to Exercise 1a Tutorial 7

134 / 9 = 14 remainder 8

The answer consists of 1,4 and 8.

1 is just the first figure of 134.

4 is the total of the first two figures 1+ 3 = 4,

and 8 is the total of all three figures 1+ 3 + 4 = 8.

Exercise 1b Tutorial 7

Divide by 9:

6) 232 =          remainder

7) 151 =          remainder

8) 303 =         remainder

9) 212 =          remainder

10) 2121 =       remainder

Answers to Exercise 1b Tutorial 7

842 / 9 = 8₁2 remainder 14 = 92 remainder 14

Actually a remainder of 9 or more is not usually
permitted because we are trying to find how
many 9's there are in 842.

Since the remainder, 14 has one more 9 with 5
left over the final answer will be 93 remainder 5

Exercise 2 Tutorial 7

Divide these by 9:

1) 771 / 9 =          remainder

2) 942 / 9 =          remainder

3) 565 / 9 =         remainder

4) 555 / 9 =        remainder

5) 2382 / 9 =       remainder

6) 7070 / 9 =         remainder

Answers

Answers to exercise 1 Tutorial 1

1) 223
2) 717
3) 495
4) 7655
5) 0124
6) 8989
7) 43
8) 943
9) 9679
10) 9962

Answers to exercise 1 tutorial 2

1) 64
2) 63
3) 72
4) 49
5) 81
6)2₁6= 36

Answers to Exercise 2 Tutorial 2

1) 8526
2) 8536
3) 7546
4) 8928
5) 8648
6) 6336
7) 9506 (we put 06 because, like all the others,
we need two figures in each part)

Answers to exercise 3 Tutorial 2

1) 10914
2) 10918
3) 10816
4) 11772
5) 12423
6) 10506 (we put 06, not 6)

Answers to Exercise 1 Tutorial 3

1) 29/30
2) 7/12
3) 20/21
4) 19/30
5) 1/20
6) 13/15

Answers to Exercise 1 Tutorial 4

1) 2025
2) 4225
3) 9025
4) 1225
5) 225

Answers to Exercise 2 Tutorial 4

1) 2021
2) 624
3) 4216
4) 221
5) 3009
6) 5621

Answers to Exercise 1 Tutorial 5

1) 294
2) 682
3) 651
4) 462
5) 672

Answers to Exercise 2a Tutorial 5

1) 987
2) 989
3) 1696
4) 1344
5) 5112

Answers to Exercise 2b Tutorial 5

6) 1792
7) 1728
8) 2232
9) 2332
10) 3456

Answers to Exercise 1 Tutorial 6

1) 473
2) 891
3) 165
4) 484
5) 121

Answers to Exercise 2 Tutorial 6

1) 968
2) 924
3) 528
4) 803
5) 616

Answers to Exercise 3 Tutorial 6

1) 1661
2) 5797
3) 3663
4) 7854
5) 9999

Answers to Exercise 1a Tutorial 7

1) 6 r 7
2) 3 r 6
3) 4 r 8
4) 5 r 8
5) 8 r 8

Answers to Exercise 1b Tutorial 7

1) 25 r 7
2) 16 r 7
3) 33 r 6
4) 23 r 5
5) 235 r 6 (we have 2, 2 + 1, 2 + 1 + 2, 2 + 1 + 2 + 1)

Answers to Exercise 2 Tutorial 7

1) 714 r15 = 84 r15 = 85 r6
2) 913 r 15 = 103 r15 = 104 r6
3) 516 r16 = 61 r16 = 62 r7
4) 510 r15 = 60 r15 = 61 r6
5) 714 r21 = 84 r21 = 86 r3
6) 2513 r15 = 263 r15 = 264 r6
7) 7714 r14 = 784 r14 = 785 r5

Tutorial 1

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Questions dealing with alkane chemistry

Problems

1. Bromination of methane is slowed down by the addition of fairly large

amounts of HBr.

a. suggest an explanation

Br . + H-CH3 ? H-Br + . CH3

?H +104 -88 = +16kcal i.e. endothermic in this direction

. CH3 + Br2 ? BrCH3 + Br .

?H = +46 -70 = -24kcal i.e. exothermic in this direction and therefore,

as soon as .CH3 forms, it reacts with Br2. Ea is low for such an

exothermic reaction.

However, as HBr is added, a competing reaction becomes possible,

the reverse of the first reaction given above:

HBr + . CH3 ? Br . + H-CH3

as ?H is ?16 kcal for this process, Ea will be very small. This reaction

removes methyl radicals and the number available for reaction with

bromine will be lessened and the reaction slows down.

b. why does HCl not react in this way

Cl . + H-CH3 ? H-Cl + . CH3

?H +104 -103 = +1kcal i.e. just endothermic in this direction. The

activation energy is +4 kcal compared with +18 for the bromination.

The activation energy for the reverse reaction is therefore +3 kcal,

higher than that for the analogous bromination reaction The methyl

radicals are also formed more readily and so are less "rare".

Note: Whereas part a) could be an exam question providing bond

dissociation energies are given, b) would never be?..

2. A mixture of H2 and Cl2 does not react at room temperature. At high

temperature or under the influence of light (of wavelength absorbed by

chlorine) a violent reaction takes place and HCl is formed. The

photochemical reaction yields as many as a million molecules of HCl for

each photon absorbed. The presence of a trace of oxygen slows down the

reaction.

a. propose a possible mechanism for this reaction

The fact that "the photochemical reaction yields as many as a million

molecules of HCl for each photon absorbed" indicates a chain reaction.

Initiation involves chlorine as light absorbed by chlorine begins the

process.

Cl2 ? 2Cl . (initiation)

Cl . + H2 ? HCl + H . (propagation)

H . + Cl2 ? HCl + Cl . (propagation)

2H . ? H2 (termination)

2Cl . ? Cl2

H . + Cl . ? HCl

b. explain why a mixture of H2 and I2 does not behave in the same

manner

I . + H2 ? HI + I . is very endothermic (+104 kcal to break H-H bond

whereas only 71 kcals are liberated in forming the H-I bond. Ea must

be > 33 kcal which is impossible. The corresponding reaction with

chlorine is only just endothermic, +1 kcal, and so its activation energy

is far lower.)

3. t-Butyl peroxide is a source of free radicals:

(CH3)3CO-OC(CH3)3 ? 2(CH3)3CO. (heat or light initiated)

A mixture of isobutane and CCl4 is stable at 130C. If a small amount of tbutyl

peroxide is added, a reaction occurs that yields 2-chloro-2-

methylpropane and chloroform. A small amount of 2-methyl-2-propanol,

(CH3)3COH, equivalent to the peroxide used, is also formed.

a. propose a mechanism which explains the formation of these products

(CH3)3CO-OC(CH3)3 ? 2(CH3)3CO.

(CH3)3CO. + (CH3)3CH ? (CH3)3COH + (CH3)3C .

(CH3)3C . + CCl4 ? (CH3)3CCl + . CCl3

. CCl3 + (CH3)3CH ? HCCl3 + (CH3)3C.

b. what is a peroxide?

a compound containing an O ? O bond

4. Why would one not prepare a Grignard reagent from bromoethanoic acid,

BrCH2CO2H?

because it contains and acidic hydrogen which would react with the

Grignard reagent as soon as it forms

5. When a mixture of methane and chlorine is irradiated, reaction begins at

once. However, when irradiation is stopped, the reaction gradually slows

down; it does not stop immediately. Why?

because it is a chain reaction and the chains continue until termination

takes place or until all of one reactant has been used up

6. The anti conformation of 1,2-dichloroethane is 1.15 kcal/mol more stable

than the gauche conformation. The two energy barriers, measured relative

to the energy of the gauche conformation, for carbon-carbon bond rotation

are 5.15 kcal/mol and 9.3 kcal/mol.

a. sketch a graph of potential energy vs. angle of rotation about the

carbon-carbon bond. Show the energy differences on your graph and

label each minimum and maximum with the appropriate conformation

of 1,2-dichloroethane.

b. Which conformation is present in the greatest amount? Explain.

The anti conformation is present in the greatest amount as it has the

lowest potential energy.

Synthesis

1. Propose a synthesis of n-hexane from each of the following:

a. CH3CH2CH2CH2CH2CH2Br

either treat with tri-n-butyl stannane or with Mg followed by water

b. CH3CH=CHCH2CH2CH3

hydrogen and Pt, Pd or Ni catalyst

c. CH3CH2CH2Br

1. Li, 2. CuI 3. CH3CH2CH2Br (Corey-House reaction)

2. Propose a synthesis of CH3D from CH4.

1. Br2/h? 2. Bu3SnD (or Mg followed by D2O)

(3)

SAMPLE FILE - IIT - CHEMISTRY
1. Choose the correct option. Only ONE option is correct.

1 mol of equimolar mixture of ferric oxalate and ferrous oxalate will require x mol of KMnO4 in

acidic medium for complete oxidation. Then x is

(a) 0.8 mol (b) 0.9 mol (c) 1.6 mol (d) 1.8 mol

2. Choose the correct option. Only ONE option is correct.

Predict the compounds A and B in the following reactions

CH CH ? O CH CH O h A

3 2 - 2 3 + 2 ¾¾n® ; C6H5CH CH3

CH3 O2, 95 ? 135º B

(a) CH3CH2 ? O-O-CH2CH3 and C6H5 C ? O ? OH

CH3

respectively

(b) CH3CH2? O ? O? CH2CH3 and C6H5 CH

CH3

CH2 ? O ? OH

respectively

(c) 2 3

|

3 H ? O ? CH CH

OOH

CH C and C6H5 C ? O ? OH

CH3

respectively

(d) No reaction and ? OOH

CH3

C H C

|

6 5 respectively

3. Choose the correct option(s), ONE or more than ONE option may be correct.

If P0 and Ps are the vapour pressure of solvent and solution respectively and N1 and N2 are

the moles of solute and solvent then :

(a)

0

s

0

(P P )

P

-

=

1 2

1

N N

N

+

(b)

0

s

P P

P

-

=

2

1

N

(c) Ps =

0

2

1 2

P .N

N +N

(d) None of these

4. Choose the correct option(s), ONE or more than ONE option may be correct.

Mg3N2 H2O A(gas) CuO B C(g) D ¾¾¾® ¾¾¾® + + H2O

C(g) can be obtained by heating

(a) (NH4)2SO4 (b) (NH4)2Cr2O7 (c) NH4NO3 (d) NH4ClO4

(4)

5. Match the items given in Column I with the items in Column II. An item of Column I can match

with more than one item from Column II. Choose the correct option from the SIX options given

below :

Match the reactions, on the left hand side with the conditions (on the right hand side), which are

most favourable for them.

Column I Column II

A. 2SO2 +O2 2SO3 + 45 K cal (P) Low temperature

B. N2 + O2 2NO ? 43.2 Kcal (Q) High pressure

C. A + B

Constant

Pressure

2C + D (R) High temperature

D. Ice Water ? X kcal (S) Addition of inert gas

(a) A-P,Q ; B-R ; C- S ; D- Q, R (b) A-P ; B-R, Q ; C- P ; D- P, S

(c) A-R ; B-P ; C- Q ; D- S (d) A-P,Q ; B-P ; C- R ; D- R, S

(e) A-S ; B-Q ; C- S ; D- P, S (f) A-R, S ; B-S ; C- S, Q ; D- Q, P

6. Read the following passage and answer the THREE questions that follows (only ONE option

correct) :

Hess's law is used to predict the heat energy released or consumed in chemical reactions. It

states that the enthalpy change is the same regardless of whether a reaction occurs in one

step or in several steps.

Below are some bond dissociation energies and enthalpies of formation.

DH (kJ/mol)

C?C 347

C?H 415

O?H 464

O=O 498

C=O 745

CºC 837

Atomisation of H - atom 217.5

DHf (kJ/mol of product)

C(s) + O2(g) ¾¾® CO2(g) ?394

H2O(l) ¾¾® H2O (g) 44

I. What is the approximate DH for the combustion of ethyne in oxygen (per mole of ethyne)?

(a) ?996 kJ/mol (b) 996 kJ/mol (c) ?1,992 kJ/mol (d) 1,992 kJ/mol

II. Estimate DH for, 2C4H10 ¾¾® C8H18 + H2 . Given the heat of atomisation of H atom is 217.5

kJ mol?1.

(a) 265.5 kJ mol?1 (b) ?48 kJ mol?1 (c) ?265.5 kJ mol?1 (d) 48.0 kJ mol?1

III. The heat of formation of liquid water is

(a) 288 kJ mol?1 (b) ?244 kJ mol?1 (c) ?288 kJ mol?1 (d) 332 kJ mol?1

plz any one tell me

ya i am also waiting

i did not recieve my score card and my counselling is on 9 july i send draft for duplicate admit card but no reply what should i do

this ios not fair thing. iit doesnt see whether u r from isc or cbse or state board . my friend was in icse board and crack iit just after passing 10+2 so dont think abt what people says

: A poor waiter in Patna has become a celebrity of sorts after his son cracked the IIT entrance exam this year. The overjoyed man gives full credit to I-Desire ? a small group of former IITians in Bihar who coach underprivileged kids and provide them study material.

Mani Bhusan Singh and his parents are getting privileged treatment in their neighbourhood after the 18-year-old cleared the Indian Institute of Technology Joint Entrance Examination (IIT-JEE). ?I am now one of the most sought after people in my workplace and also my neighbourhood,? says a proud Lallan Singh, Mani Bhusan?s father.

?I got to know about I-Desire through a friend. They agreed to help my son prepare for IIT-JEE. It wouldn?t have been possible if I-Desire had not helped us by providing coaching, study material and a proper guide,? said Lallan Singh, a waiter in Hotel Maurya in Bihar.

Lallan Singh, in his 40s, said his son could not appear for CBSE exams in Class 10 and 12 due to financial constraints. ?My son wanted to appear for the CBSE exams but due to lack of money I forced him to appear for the Bihar Examination Board exams,? said Lallan Singh, who lives in a one-room rented house in Mandiri, a densely populated lower middle class area.

There are 15 other students from underprivileged families here who have made it to IIT thanks to I-Desire. The former IITians have been funding I-Desire, which provides free coaching for a year and counselling by IITians, explained Grijesh Jha, coordinator of I-Desire and a former student of IIT Kanpur. ?As many as 16 of the 35 boys selected by us cracked the IIT exam this year,? Mr Jha said.

I-Desire, a brainchild of Sushil Kumar, another IIT Kanpur product, was set up three years ago to provide help and support to IIT aspirants from the weaker sections of the society in Bihar. Mr Kumar, who hails from Patna, is currently working for a multinational company in Chicago, said Mr Jha.

The project was started in memory of murdered engineer Satyendra Dubey, who blew the whistle on corruption on the national highways project in Bihar. IIT engineer Mr Dubey, who worked on the Bihar section of the national highways project in Gaya district, was allegedly killed for writing to the Prime Minister?s Office on corruption in the project. He was shot dead in Gaya town, about 100 km from Patna, on November 27, 2003.

English chemistry mnemonics

Oxidation and reduction

Periodic table

Hydrogen Bonds

Dicarboxilic Acids (electrodes)

Oxidation and Reduction (electrons)

Sentences to remember the elements

Mnemonics:

1)

2)

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