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Depression in freezing POINT (ΔTf) = KF x m

0.186 = 1.86 x m

m = 0.1

Now as we have got the molality, m= 0.1

ELEVATION in boiling point (ΔTb) = Kb x m

→ ΔTb = 0.512 x 0.1

→ ΔTb = 0.0512

Therefore, the elevation in Boiling point is 0.0512.

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(sodium bicarbonate or BAKING SODA) NaHCO3 + HC2H3O2 (vinegar or acetic acid). products are water, carbon DIOXIDE, and sodium ACETATE (NaC2H3O2).

It has the ABILITY tooxidize other SUBSTANCES — in other WORDS to cause them to lose electrons....
that the reason why it acts as an OXIDIZING agent....
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Hydrogen gas gives a POP SOUND when it enters base SOLUTIONS.(Soap is a basic PRODUCT).
The pop sound represents the bubbles.

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1. To separate different gases from LIQUID AIR.
2.to separate various useful FRACTIONS such as gasoline, KEROSENE oil, diary oil, etc. from crude oil in petroleum industry.



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The MATHEMATICAL product of the separation of the ENDS of a dipole and the MAGNITUDE of the charges.

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On heating it LOOSES water MOLECULES and becomes anhydrous ferrous sulphate . 2. Colour changes from light green to white . Ferrous sulphatecrystals which are in green colour decomposes to give Ferric oxide , sulphur DIOXIDE and sulphur trioxide gas.

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↘Gold foil was USED by Ernest Rutherford for his ALPHA PARTICLES SCATTERING experiments as Gold is an UN reactive ( noble metal ) . It is malleable and hence it's thin foil allowed easy deflection of the alpha particles.

This might not be possible with the other heavy as well as light metals .

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The hydrides of group 16 have tetrahedral geometry. With TWO lone pairs of the ELEMENTS. There is an increase in the bond angle as we move down the group because

-The ELECTRON negativity of elements decrease as we move down.

-The lone PAIR - bond-pair repulsion is stronger is than lone pair-lone pair or bond pair bond-pair interaction.

-But as the electronegativity decreases, the repulsion also decreases which RESULTS in a decrease in bond angle.

A) baking soda is a mild BASE which can dilute the excess ACID formed during indigestion. Hence it acts as an antacids.
B)Baking soda when heated ,forms carbon-dioxide. This co2 surrounds the combustible SUBSTANCE and cuts the oxygen supply. Finally the fire gets extinguished.
C)when baking soda is heated, co2 is formed ,which is responsible for making the cake FLUFFY and soft as well.

Glycerol reacts with OXALIC acid in two ways. Formation of formic acid. When glycerol is HEATED with crystalline oxalic acid at 100 - 110°C, formic acid is PRODUCED. ... When heated with oxalic acid at 260°C, allyl ALCOHOL is formed.

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Answer:

Kc=[SO3]2[SO2]2⋅[O2]

Explanation:

For a CHEMICAL equilibrium, the equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the PRODUCTS and the product of the equilibrium concentrations of the reactants, all raised to the power of their respective STOICHIOMETRIC coefficients.

It will INCREASE DEFINITELY

So DILUTE hydrochloric acid is REACTING here

HCl(DIL.)+NA2CO3--->NaCl+H20+Co2

ELECTRON gain ENTHALPY is the ENERGY released when one mole of electrons are added to the isolated gaseous ATOMS of an element.

Electron gain enthalpy is said to be positive an electron is added to an element

And, it is negative when an electron is removed from the element

Since the ELECTRONIC configuration of nitrogen is 2s2 2p3, it has a half filled p shell which is a stable configuration. Thus, it has a relatively low tendency to gain or lose electrons and its electron gain enthalpy is zero.

Because there is no any other places for nitrogen
therefore it ATTACK on terminal position and after HYDROLYSIS ony PRIMARY amine is OBTAINED

Hey there!!!

Answer :-

It represents the NUMBER of atoms in 1 gram atom of an element or the number of molecules in one gram mole of a compound if we divide the atomic MASS of an element by actual mass of its atom it's value is 6.022 into 10 raise to POWER 23 .



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☺❤☺❤☺HEY MATE YOUR ANSWER IS HERE. .....☺❤☺❤☺

Mendeleev NAMED the undiscovered elements using the Sanskrit word Eka (meaning one) as a prefix, with the name of the preceding element in the same group. For example, GALLIUM was not DISCOVERED in Mendeleev’s TIME. 

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This is your ANSWER to your QUESTIONS

Diffusion (and effusion),in fundamental terms, follow Graham's law, which can be stated as:

Rate1/Rate2 = √(M2/M1)

where:

Rate1 and Rate2 = rates of diffusion of the two gases (often in terms of moles per

unit time)

M1 and M2 = molar masses of the the two species (g/mol)

Let's call hydrogen species 1 here, and oxygen species 2.

We are told 4 g of O2 diffuse in a certain time. 4 g of O2 = 1/8 mol, since the molar mass of O2 is 32 g/mol. So the rate at which O2 is DIFFUSING is (1/8) mol in t SECONDS, or ((1/8)/t) mol/s

So, for quantities to put into Graham's law, we have:

M1 = 2 g/mol (for hydrogen)

M2 = 32 g/mol (for oxygen)

Rate1 = UNKNOWN (we are trying to solve for this)

Rate2 = ((1/8)/t) mol/s

So, using Graham's law, we can now solve for the rate of hydrogen flow in mol/s.

(Rate1/Rate2) = √(M2/M1) --> (Rate1/[(1/8)/t)]) = √(32/2) = √16 = 4

Thus, hydrogen diffuses four times faster than oxygen, on a mole/second basis.

So, Rate1 = [(1/8)/t]*4 = (4/8)/t = (1/2)/t

Now, the diffusion of hydrogen is said to occur over the same period as the diffusion of oxygen, so Rate1 is measured over the same time. Thus:

Rate1 = (mol of hydrogen diffusing)/t,where t is the same t as above (for Rate2).

So, combining the LAST two equations, we GET:

(number of moles of hydrogen diffusing)/t= (1/2)/t

Thus, the number of moles of hydrogen diffusing is 1/2 mol. 1/2 mol of molecular

hydrogen has a mass of 1/2*(2 g/mol) = 1 g.

Thus, 1 g of hydrogen would diffuse in the same time as 4 g of oxygen. This may

seem counter-intuitive, but remember that Graham's law relates molar diusion

rates, not mass diffusion rates. And, even though the hydrogen diffuses

faster, it is also 16 times lighter than oxygen. So, in mass terms, only half as much mass of hydrogen diffuses in the same time.

No. of MOLES =GIVEN weight÷molecular weight.
So,
ACC to question
n=56÷20
=2.8

Hey mate is your ANSWER

c=c BOND is more stronger than c-c bond as it FORMS a covalent bond i.e. formed by mutual SHARING of electrons.

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There are 56 ELEMENT were KNOWN at NEWLAND's TIME

Latent heat of vaporization is a PHYSICAL PROPERTY of a substance. ... When a material in LIQUID state is given energy, it CHANGES its phase from liquid to vapor; the energy ABSORBED in this process is called heat of vaporization. The heat of vaporization of water is about 2,260 kJ/kg, which is equal to 40.8 kJ/mol.

Wh N oxidation and REDUCTION both are TAKING place in one reaction than its CALLED REDOX reaction PbO is being reduced in above reaction because it losses oxygen in the rection

Hi Nishith.......

here is ur ans.....

MOLECULAR FORMULA of methane is CH₄

and molecular formula of ethane is CH₃-CH₃ or C₂H₆.

hope U got it.

With LOVE. All the best.

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Dehydrated means in simple WORDS is WATER LOSS

Compound A with molecular formula C8H8O give positive DNP and iodoform test it neither reduce tollens REAGENT nor does it decolourise BROMINE water white.The structure of A is BENZOPHENONE i.e PhCOCH3.

As it is a ketone, it can not reduce tollen's reagent to give silver mirror. It also does not decolourise bromine water as double BOND of phenyl RING is involved in reaction with bromine water.Hence this compound A is nothing but a ketone.

It MIGHT be 3RD ORDER reaction
it will CONFIRM only by EXPERIMENTS

Hey mate there is your ANSWER :

Thermite reaction : The process of igniting aluminium and ferric oxide is called thermite reaction. This is an exothermic reaction.

Fe2O3+Al=====>Al2O3 +2FE +HEAT

Answer : The half life of the reaction is, 23.1 minutes.

Solution : GIVEN,

RATE of first ORDER reaction =

Concentration of reactant = 0.5 m = 0.5 mole/L

• For first order reaction,

The expression for rate of reaction is,

Now put all the given values in this formula, we get the value for rate constant.

The formula for half life for first order reaction is,

Now put the value of K, we get the value of half life.

Therefore, the half life of the reaction is, 23.1 minutes.

NaCn is use for leaching of Ag ORE in the presence of air from which the silver is OBTAINED LATER by replacement.

4Ag + 8CN- + 2H2O + O2 → 4 [Ag(CN)2-] + 4OH-

Yes when magnesium RIBBON is BURNED it FORMS magnesium OXIDE when this magnesium oxide is dissolve with WATER it forms magnesium hydroxide

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Explanation:

There are TWO main EXCEPTIONS to electron configuration: CHROMIUM and copper. In these cases, a completely FULL or half full d sub-level is more stable than a partially filled d sub-level, so an electron from the 4s ORBITAL is excited and rises to a 3d orbital.

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Your QUESTION is in this way
Calculate the ENTROPY change for the conversion of 1 GM of ice to WATER at 273 K and one atm pressure. The enthalpy of fusion of ice is 6.025 kJ/mol

ΔSfusion=ΔHfusionTfΔSfusion=ΔHfusionTf=6025J/mol273K=6025J/mol273K=22.1J/K−mol=22.1J/K−mol

Now, entropy change for conversion of 1 mol, i.e, 18g of ice into water at 273K=22.1J/K273K=22.1J/K

So, Entropy change for converting 1g of ice into water = 22.1J/K1822.1J/K18=1.227J/K

The FORMULA is
C2H50H

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(1) By electrophoresis: In electrophoresis the colloidal particles move towards oppositely charged electrode. When these come in contact with the electrode for long these are discharged and precipitated.

(2) By mixing two oppositely charged sols : When oppositely charged sols are mixed in almost equal proportions, their charges are neutralised. Both sols may be partially or completely precipitated as the mixing of ferric hydroxide (+ve sol) and arsenious sulphide (–ve sol) bring them in precipitated form. This type of coagulation is called mutual coagulation or meteral coagulation.

(3) By boiling: When a sol is boiled, the adsorbed layer is disturbed due to increased collisions with the molecules of dispersion medium. This REDUCES the charge on the particles and ultimately they settle down to form a precipitate.

(4) By persistent dialysis: On prolonged dialysis, the traces of the electrolyte present in the sol are removed almost completely and the colloids become unstable.

(5) By addition of electrolytes: The particles of the dispersed phase i.e., colloids bear some charge. When an electrolyte is added to sol, the colloidal particles take up ions carrying opposite charge from the electrolyte. As a RESULT, their charge gets neutralised and this causes the uncharged, particles to come closer and to get coagulated or precipitated. For example, if solution is added to sol the ions are attracted by the negatively charged sol particles and their charge gets neutralised. This lead to coagulation.

(6) Hardy schulze rule : The coagulation capacity of different electrolytes is different. It depends upon the valency of the active ion are called flocculating ion, which is the ion carrying charge opposite to the charge on the colloidal particles. “According to Hardy Schulze rule, greater the valency of the active ion or flocculating ion, greater will be its coagulating power”thus, Hardy Schulze law state:

(i) The ions carrying the charge opposite to that of sol particles are effective in causing coagulation of the sol.

(ii) Coagulating power of an electrolyte is DIRECTLY proportional to the valency of the active ions (ions causing coagulation).

For example to coagulate negative sol of , the coagulation power of different cations has been found to decrease in the order as,

Similarly, to coagulate a positive sol such as , the coagulating power of different anions has been found to decrease in the order :

(7) Coagulation or flocculation value

“The minimum concentration of an electrolyte which is required to cause the coagulation or flocculation of a sol is known as flocculation value.”

or

“The NUMBER of millimoles of an electrolyte required to bring about the coagulation of one litre of a colloidal solution is called its flocculation value.”

Coagulation value or flocculating value

(8) Coagulation of lyophilic sols

(i) There are two FACTORS which are responsible for the stability of lyophilic sols.

(ii) These factors are the charge and solvation of the colloidal particles.

(iii) When these two factors are removed, a lyophilic sol can be coagulated.

(iv) This is done (i) by adding electrolyte (ii) and by adding suitable solvent.

(v) When solvent such as alcohol and acetone are added to hydrophilic sols the dehydration of dispersed phase occurs. Under this condition a small quantity of electrolyte can bring about coagulation.

Carbon has catenation tendancy due to which it can FORM LONG CHAINS .
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Given, V1 =2500 cm3
                   P1 =760mm [At S.T.P]

  It is said that Hydrogen is subjected to 2.5 times increase in pressure.
  Thus final pressure , P2 of the gas is = 2.5 x P1
                                                          = 2.5 x 760
                                                     P2  = 1900 mm

We have to find out V2  [final VOLUME] which the gas occupies
From Boyle's Law, At constant temperature

                                         P1V1 = P2V2

​                                       Thus , V2  =    P1V1P2
                                           V2  =   2500×7601900
                                        
                                           V2 = 1000cm3


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A total of four quantum NUMBERS are used to DESCRIBE completely the movement and trajectories of each electron WITHIN an atom. The combination of all quantum numbers of all electrons in an atom is described by a wave function that complies with the Schrödinger equation.

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OXYGEN is a smaller ATOM and is more electronegative, so H2O WOULD be more polarized and have smaller BOND angle. Thus, H2O has higher DIPOLE moment.

Liquor Ammoniae Fortis is a STRONG Solution of Ammonia. ... Nh3 - An aqueous solution CONTAINING 33% of ammonium, NH3. A COLORLESS LIQUID with a PUNGENT, suffocating odor.