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The GENERAL name of fluid FORM the MATTER is liquid.

Let there are x and y moles of CO2 and CO. now find tot MASS and volm wrt x and y and EQUATE it to given quantities.

It is KNOWN as SOLID...

The important condition necessary for the formation of covalent BONDS are:-
1) the elements must have high electronegativity values..
This should not be CONFUSED between electronegativity difference values..
As the covalent bond is formed by MUTUAL sharing of electrons ,hence the shared species must have a lesser electronegativity difference among Themselves..
like CH4 (methane) ..the electronegativity difference values between C and H is very less as compared to HCl(ioniccompound) where H and Cl share greater electronegativity difference...
2) the ionisation energy of the mutually shared elements must be high..

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Fluid system in which LIQUID droplets are dispersed in a liquid. ... An emulsion is a mixture of TWO or more liquids that are normally IMMISCIBLE (unmixable or unblendable).Emulsions are part of a more general class of two-phase systems of MATTER calledcolloids.

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An aerosol is a suspension of fine solid particles or liquid droplets, in air or another gas. Aerosols can be natural or [HUMAN impact on the environment. ... Examples of natural aerosols are FOG, dust, forest exudates and GEYSER steam. Examples of anthropogenic aerosols are haze, particulate air pollutants and smoke.


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The delocalised electrons in the metalare FREE to move throughout the entire structure. When attached to an electrical power source, this means THEMETAL can conduct electricity. ...Metals are therefore said to be malleable (easily beaten into SHAPE), and DUCTILE (easily pulled out into wires).


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MERCURY is the only one in liquid state at room temperature. It's used inthermometers because it has high coefficient of expansion. Hence, the slightest CHANGE in temperature is notable when it's used in athermometer. It also has a high BOILING POINT which makes it very suitable to measure higher temperatures.


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When it reacts with NANH2 it gets reduced to C2H4. It reacts with CH3I to form CH3CH=CH2 which upon the last reaction FORMS acetone CH3COCH3.

For LAST line of lyman series N1 = 1 , N2 = infinte

1/LAMDA = RZ^2 (1/n1^2 - 1/n2^2)

1 / lamda = R (1/1 - 0)
= R
= 109700cm^-1

v = C/lamda
= C * R
= 3 * 10^8 * 109700
= 3.29 *10^15 / sec

Element - helium , nitrogen , HYDROGEN
compound - WATER , calcium oxide , carbon dioxide , sugar , salt
mixture - air , SEA water , salt sol. , 22 KARAT GOLD , brass , smoke

Resonance is a method in the Valence Bond Theory of bonding to account for and to describe the delocalization of electrons within MOLECULES. Resonating structures are employed when one structure for a single MOLECULE cannot fully describe the bonding that takes place between neighboring atoms. Resonance helps to represent the possible ion intermediates by having the EXTRA electron or electrons delocalized within the molecules. A molecule that has several resonance structures is more stable than a molecule that has less resonance structures.

The different structures by which the molecule is REPRESENTED are called resonating structures or canonical forms.

Some of the examples of resonance structures are as follows:

In ozone, O3, there are TWO equivalent resonance structures.

In benzene, we find resonance forms by the curved arrow formalism, which is a way of depicting the “movement” of electrons.

1) O2
Will be the CORRECT OPTION

OPTION 3
1/2 RT
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Packing fraction is the percentage of the total space OCCUPIED by the particles

In HEXAGONAL close packing, the packing fraction is 74%

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The NORMAL BLOOD PH is TIGHTLY REGULATED between 7.35 and 7.45.

A meso compound or meso ISOMER is a stereoisomer with an identical or superposable MIRROR image i.e., a non-optically active member of a set of stereoisomers, at least TWO of which are optically active. 

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The angular momentum quantum NUMBER, l, (also referred to as the secondary quantum number or azimuthal quantum number) DESCRIBES the SHAPE of the orbital that an electron occupies. The lowest possible value of l is 0, and its highest possible value, depending on the PRINCIPAL quantum number, is n - 1.
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From the given ratio we can get the masses of the two compounds.

From the ratio the mass of CH₄ will be :

5/7 × 280 = 200 g

The mass of C₂H₆ will be :

2/7 × 280 = 80 g

Equation for COMBUSTION of both is :

CH₄ + 2O₂ —> CO₂ + 2 H₂O

The moles of CH₄ are :

200 / 16 = 12.5 moles.

The mole ratio of CH₄ to CO₂ is 1 : 1 and hence moles of CO₂ are 12.5 moles.

C₂H₆ + 3.5O₂ —> 2CO₂ (g) + 3H₂O

Moles of C₂H₆ are :

80/30 = 2.667 moles.

The mole ratio is 1 : 2

Moles of CO₂ will be :

2 × 2.667 = 5.334 moles.

Total moles :

5.334 + 12.5 = 17.834 moles

The average atomic mass of nitrogen is14.0067. The atomic masses of TWO STABLE isotopes of nitrogen, nitrogen-14 and nitrogen-15, are 14.003074002 and 15.00010897, respectively. Use this INFORMATION to determine the percent ABUNDANCE of each nitrogen ISOTOPE.

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350.15kelvin is the CORRECT ANSWER

When acis reacts to metals it evolve gas H2
let me GIVE you example...

2HCL +Mg--->MgCl2+H2

The MOLARITY of the GIVEN SOLUTION is 1M

The ONE with LOWER bp bcoz it requires LESS ENERGY to vapourize than that of HIGHER bp

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pH Formula. pH is a measure of the activity of hydrogen ions (H+) in a solution and therefore its acidity or alkalinity. ... The NUMBER of places after the DECIMAL point in the pHequals the number of significant FIGURES reported in the HYDRONIUM ion concentration.

PH FORMULA :- pH = -LOG[H3O+]

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1.Helium
2.Neon
3.Argon
4. Krypton
5.Xenon
6.Radon
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What is the QUESTION?

Zn+2HCl - - - - >ZNCL2 +H2

When GLUCOSE is react with HI for a LONG time then n-hexane is formed..

The NUMBER is 8.




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Decomposition of FERROUS SULPHATE into FERRIC oxide,sulphur DIOXIDE,sulphur trioxide.

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Volume of oxygen taken = 30 ml
Volume of unused oxygen = 15 ml
Volume of O2 USED = Volume of O2 added – Volume of O2 left
= 30 – 15 = 15 ml
Volume of CO2 produced
= Volume of gaseous mixture after explosion – Volume of unused oxygen
Or Volume of CO2produced = 25 – 15 = 10 ml
Volume of HYDROCARBON = 5 m;
General equation for combustion of a hydrocarbon is as follows –
CxHy + (x + y/4)
O2 → xCO2 +y/4 H2O
(Hydrocarbon)
5 ml 5(x + y/4)ml 5 x
∴ Volume of CO2 produced = 5X, Since Volume of CO2 = 10 ml
∴ 5x = ⇒ x = 2, Volume of O2 used = ml
∴ 5 (x + y/4) = 15 ⇒ x + y/4 = 3
⇒ 2 + y/4 = 3 (∵ x = 2)
⇒ 8 + y = 12 ∴ y = 4