Ans. antilog 0.1170

1) Separate characteristic and mantessa part 

 characteristic part is 0

 mantissa part is .1170

2) Use antilog table to find corrosponding value for mantissa:

open the antilog table and find the row number beginning with .11, then the column for 7.

3) Find the value from mean difference column :

use the row number beginning with .11, but find the column for fourth digit 0.

4) add both values you get in step 2 and 3.

5) inserting decimal point : after the number of digits that corresponds to the characteristic plus 1. add after 0+1 = 1 digit

It is not the Fructose molecule that acts as a reducing sugar. 
It is the solution.
In the solution of Fructose some molecules get converted to Glucose and that is how it exists.
Fructose and Glucose are interconvertible structures .
Therefore in the Fructose solution it is the Glucose molecule that acts as a reducing sugar

Ans. The melting point is a good measure of how much energy has to be inserted into the system to break the intermolecular bonds. Mn and Tc still forms these bonds, since these are present as a solid at some particular temperature.

However, the bonds they form are proportionally weaker because the electrons don't participate as strongly in the bonding. It's not that the density of electrons is lower in the delocalized system but that these electrons intrinsically have a lower stabilization resulting from the delocalization as compared to some other metal elements.

Thus, the bonds are weaker, the delocalized solid system is not as preferred, and less energy is necessary to melt the metal.

Ans. Allotropy is the property of some chemical elements to be able to take two or more different forms, where the atoms are arranged differently by chemical bonds<wbr />.

Allotropy is the property of some chemical elements to exist in two or more different forms. For example Carbon has three allotropes namely Diamond, Graphite and Buckminster Fullerene.

They are all made of Carbon atoms but because of their different structures, each one of them ha diffrerent  physical properties.

yes isopropyl alcohol gives iodoform test. 

Reaction --  CH_3-CHOH-CH_{3 +} I_{2 +} NaOH  --------------->  CH₃COONa + 

                           CHI₃ +NaI + water 

Stick to NCERT for everything. Physical is easy, practice numericals (specially NCERT questions)  learn definitions and the read NCERT word to word.
Organic, something people fear. Don't fear it. Make a list of all the important reactions you come across and revise them daily! Names of the compounds, reagents and everything. Learn them by heart. Practice as many questions as you can. Refer to previous year and sample paper questions. Redo them until you're absolutely sure of them. Understand and then learn, will benefit you a lot. Make sure you know all the name reactions, it'll fetch you 3-4 marks easily. Make sure that you do all the NCERT questions thoroughly, be it examples or back exercise questions.
Chemistry exam is usually simple and direct. So don't worry much. 
Don't miss the last chapters, they are very scoring if done properly.
And pdf blocks. Do NCERT religiously! Every line is important. Try to answer all questions in the language used in NCERT, specially for reasoning questions.

Along with the below answer you can sometimes just easily determine the order of reaction just by the unit/dimension of the rate of reaction. 

ZERO ORDER:
<font size="+2">rate = k[A]⁰</font> 
<font size="+2"> M/t = k</font> 

<font size="+2">K units:  M/s, M/min, M/hr, etc.</font>

FIRST ORDER: 
<font size="+2">rate =K[A]</font> 
<font size="+2"> M/t =K M</font> 
<font size="+2">  Kunits: s^-1, min^-1, hr^-1, etc.</font>

SECOND ORDER:
<font size="+2">rate =K[A]²      rate =K[A][B]</font> 
<font size="+2"> M/t =KM²</font> 

<font size="+2">Kunits: M^-1s^-1, M^-1min^-1, M^-1hr^-1, etc.</font>

THIRD ORDER: 
<font size="+2">rate =K[A]³      rate =K[A]²[B]     rate =K[A][B][C]</font> 
<font size="+2"> M/t =K M³</font> 

<font size="+2">Kunits: M^-2s^-1, M^-2min^-1, M^-2hr^-1, etc.</font>

N ORDER: 
<font size="+2">rate =K[A]ⁿ</font> 
<font size="+2"> M/t =K Mⁿ</font> 

<font size="+2">Kunits: M^-(n-1)s^-1</font>

Ans. The order of a chemical reaction is defined as the sum of the powers of the concentration of the reactants in the rate equation of that particular chemical reaction. A reaction can also have a reaction order with respect to a reactant if the rate is not simply proportional to some power of the concentration of that reactant.

Consider a general reaction:

$aA + bB \to cC + dD$

The rate expression for this reaction is (say) :  $-r_a = k{A}^x{B}^y$

Here,  x and y indicate how sensitive the rate is to the change in concentration of A and B.

Hence, the order of this reaction is (x + y).

Note that the x and y may not respectively correspond to the stoichiometric coefficients of A and B.

Reducing agent is a substance that reduces other and itself gets oxidised.mostly alkali metal hydrides are used as reducing agent as they can easily loose electrons and shows oxidation.

A substance which undergoes oxidation but reduces others is known as reducing agent or reductant. In other words a reducing agent is an electron loser.

so by this we can say that reducing agent undergoes oxidation.

it happens when an atom or compound loses one or more electrons. Some elements lose electrons more easily than others. These elements are said to be easily oxidized. Generally speaking, metals including sodium, magnesium, and iron are easily oxidized.

Secondary and tertiary amines do not for isocyanides when made to undergo carbylamine reaction. Since they have more than one alkyl group which hinders the approach of -NH₂ group.

Carbylamine reaction is a test to distinguish primary amine from secondary and tertiary amines.Now when we check the mechanism of this reaction, we find that we need to protons in the mechanism which are provided by primary amines.Seconary and tertiary amines do not give two protons , so they do not show carbylamine test.

Carbylamine test leads to formation of alkyl cyanide and this is characteristic reaction for primary amines.
Secondary and tertiary amines do not allow the formation of cyanide due to insufficient bonds for formation of triple bonds between carbon and nitrogen.
So this reaction is not given by secondary and tertiary amines.

Peptusation is the process of converting precipitate into colloidal solution with the help of dissolving the precipitate in dispersion medium in presence of electrolyte

Peptisation is the process which is the primary cause for the formation of colloidal particles in dispersion medium. It requires an electrolyte known as peptizing agent for the completion of the process. It plays a vital role for understanding precipitation reactions that occur in aqueous solution. It has a number of uses:

1. It can be used in nano particle synthesis.
2. It can be used to split large grouping into smaller particles.
3. It is used for synthesizing Titania.
4. It helps in filtrating colloidal solution.

Ans. Peptization is the process responsible for the formation of stable dispersion of colloidal particles in dispersion medium. In other words it may be defined as a process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of small amount of electrolyte.

HVZ Reaction also known as Hell Volhard Zelinsky reaction is a reaction to convert carboxylic acid with alpha hydrogen to alpha haloacid.

The reagent used are Br2,PBr3 and H2O.the first step involves conversion of carboxylic acid to acid bromide and the second step an alpha hydrogen is replaced with bromine atom (halogen atom).

Ans.Treatment with bromine and a catalytic amount of phosphorus leads to the selective α-bromination of carboxylic acids.

Mechanism of the Hell-Volhard-Zelinsky Reaction

Phosphorus reacts with bromine to give phosphorus tribromide, and in the first step this converts the carboxylic acid into an acyl bromide.

An acyl bromide can readily exist in the enol form, and this tautomer is rapidly brominated at the α-carbon. The monobrominated compound is much less nucleophilic, so the reaction stops at this stage. This acyl intermediate compound can undergo bromide exchange with unreacted carboxylic acid via the anhydride, which allows the catalytic cycle to continue until the conversion is complete.

The HVS (Hell-Volhard-Zelinsky) reaction is an organic reaction used to convert a carboxylic acid with an α-hydrogen and a halogen, to an α-halo carboxylic acid, using a phosphorous catalyst and water.

The mechanism begins with the reaction of the carbonyl oxygen with phosphorous trihalide to form a P–O bond and release a halide anion. The halide then attacks the carbonyl to form an intermediate which rearranges to release an acyl chloride, an acid molecule, and a phosphine oxide.

The acyl chloride then tautomerizes to the enol form which subsequently attacks the halogen molecule to form an α-halo acyl halide. Water hydrolysis yields the final α-halo carboxylic acid product.

Because it is stable compared to LiAlH4, it can be used in aqueous solutions and measured in open air where as LiAlH4 will react violently to moisture and water.

Sodium  borohydride  is  a  much  less hazardous  reagent  than  lithium aluminum  hydride. The  greater selectivity and  safety  of NaBH4 make  it  the preferred  reagent  in applications with Ketones.

Blister copper consists of about 2% of impurities consisting of cliver, Glod, Zinc, Nicket etc. It is mostly purified by electrolytic method. A block of impure copper is anode, a strip of pure copper is cathode while solution of CuSO₄ containing dil H₂SO₄ is e;ectrolyte. On passing current, impure copper dissolves and equivalent amount of pure copper is deposited at cathode. Impurities are collected below anode as anode mud.

             CuSO₄ (aq)      →      Cu⁺⁺ + SO^—₄

            At cathode: Cu⁺⁺ + 2e   →     Cu

            At anode:        Cu         →      Cu^{++ + 2e}

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Cu⁺ ion oxidizes and reduces simultaneously in aqueous solution to give Cu²⁺ and Cu⁺ ions. Therefore Cu⁺ ion is unstable in aqueous solution and it disproportionates readily to Cu²⁺ and Cu⁺ Ions.

In an aqueous medium, Cu2+ is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ion in an aqueous solution is unstable. It disproportionates to give Cu2+ and Cu.

Stability in aqueous conditions depends on hydration energy of ions when they bond to molecules. Cu2+ ions has greater charge density than cu+ ions and hence forms much stronger bonds releasing more energy.

Cu+ oxidises result in aqueous solution to Cu2+ and Cu. Cu2+ is more stable than Cu+ in aqueous solution as it has high hydration energy than Cu+.

Ans. Actinides exhibits larger oxidation states than lanthanide because of very small energy gap between 5f, 6d and 7s subshells . The energies are decided on the basis of (n+l) rule. The (n+l) values of the tree orbitals are as under:

5 f = 5 + 3 = 8

6 d = 6 + 2 = 8

7 s = 7 + 1 = 8

All of the values comes out to be same . Hence they have same energy.

Covalent Bond: When two atoms having similar or very low electronegativity difference, react together, they form a covalent bond by sharing electrons. Both atoms can obtain the noble gas electronic configuration by sharing electrons in this way. Molecule is the product resulted by the formation of covalent bonds between atoms. e.g. Cl₂ O₂

Coordination Bond : This is also a type of covalent bond where the two electrons in the bond are only donated by a single atom. This is also known as a dative bond. This type of covalent bonds is formed when a Lewis base donates an electron pair to a Lewis acid. e.g. BF₃

benzyl chloride is boiled under a reflux condenser with an excess of a 10 %, potassium carbonate solution for 12 hours. When the reaction is complete and the smell of the benzyl chloride  has disappeared the reaction mixture is cooled and extracted with Ether. The Ether extract is dried with potassium carbonate or sodium sulfate, and then removed by evaporation. The benzyl alcohol is purified by distillation collecting fraction which passes over at 206° C and forms a colorless liquid.

Benzyl alcohol is prepared by the hydrolysis of benzyl chloride using sodium hydroxide.C₆H₅CH₂Cl + NaOH → C₆H₅CH₂OH + NaCl

It can also be prepared by a Grignard Reactionby reacting phenylmagnesium Bromide  (C₆H₅MgBr) with formaldehyde, followed by acidification.

A carboxylic acid can be represented by resonance structure and is stabilized (as shown below)

Fig. Resonance structure of carboxylate ion.

A carboxylate anion is more stabilized by resonance. Though phenoxide ion is resonance stabilized its stability is less compared to a carboxylate ion. This is because a carboxylate ion has two equivalent structures, which are equally favored when compared to the five resonance structure of phenoxide ion (refer the resonating structures as mentioned above), all of which are not equivalent. The pKa of carboxylic acid for example acetic acid is 4.5 whereas phenol has pKa of 9.9, which shows that a carboxylic acid is stronger acid than phenol.

Carboxylic acids dissociatein water to form carboxylate ion and hydronium ion. The carboxylate ion formed is stabilized through resonance by effective delocalization of the negative charge.carboxylate ion, the conjugate base of carboxylic acid is stabilized by two equivalent resonance structures in which the negative charge is effectively delocalized between two more electronegative oxygen atoms.

On the other hand in case of phenols,negative charge is less effectively delocalized over one oxygen atom and less electronegative carbon atoms in phenoxide ion. Therefore, the carboxylate ion exhibits higher stability in comparison to phenoxide ion. Hence, the carboxylic acids are more acidic than phenols.

Nitrogen is smaller in size hence it forms multiple bonds with other nitrogen and exists as diatomic molecule. Whereas phosphorus atomic size is more so it can form single bond with other phosphorus atoms. Hence it exists as tetra atomic molecule.

Due to larger atomic size PP is unable to form pi bonds and so it is tetra-atomic in which each PP atom is linked with 3 other PP atoms by 3 sigma bond. But, due to smaller atomic size NN forms 1 sigma and 2 pi bonds i.e. triple bonds with other NN atom and exists as diatomic molecule. 

Sea water contains a lot of electrolytes. River contains colloids of sand and clay. When they meet the electrolytes neutralise the charge on colloidal particles and result in the precipitation of sand, clay etc. Thus resulting in a delta formation .

Deltas are landforms or wetlands that form at the mouth of rivers as they empty their water and sediment into another larger body of water, such as an ocean, sea, or lake, or even another river. ... The finest material is deposited beyond the river's mouth. This material is called alluvium or silt.

Role of cryolite in metallurgy of alumina:

• it functions as a solvent to alumina
• it lowers the melting point of alumina from 2323K to1140K
• reduces the cost of process
• melt of alumina and cryolite conducts electricity.

One of the methods of estimation of ozone is by iodometric method.Here ozone is allowed to react with KI.Ozone being an oxidising agent oxidises KI to iodine.The liberated iodine is reacted with sodium thiosulphate solution.From the amount of thiosulphate solution consumed in the reaction estimations are then done.

The methyl group slightly takes away electron density from the oxygen atom, unlike the hydrogen in water, which actually contributes more electron distribution.When considering the acidity of an compound, consider these four elements:

1.  Attached electronegative atom
2.  Resonance structures
3.  Size of molecule
4.  Induction (proximity to other electron-withdrawing groups)
4th is the reason why methanol is more acidic than water, but as you probably know, it doesn't contribute much

Due to the +I group of CH3 group, the bond in CH3OH becomes weaker and breaks into CH3O- where as bond energy in H2O is more.Hence H2O is more basic than methanol.Methanol is acidic than water because its conjugate base methoxide is weaker than waters conjugate base hydroxide.

Due to +Ieffect of methyl group