Enthalpy of vaporization is the amount of energy that must be supplied to the liquid in order to change the liquid into a gas. Enthalpy is also regarded as boiling temperature of the liquid. 

Acetone does not have hydrogen bond, and thus intermolecular forces are comparatively weaker which makes it boil/evaporates fast, thus reducing the molar enthalpy of vaporization. Also, Acetone does not polar O-H bond contributing to low enthalpy. 

Water has non-polar region and also has strong hydrogen bond. Water has highly polar O-H bond which makes water boil at relatively higher temperature than Acetone thus has enthalpy greater than that of Acetone. 
 

Gud eve dono ko??

Dimensional Formulas for Physical Quantities Physical quantityUnitDimensional formulaAcceleration or acceleration due to gravity ms–2 ,LT–2,Angle (arc/radius)rad,MoLoTo ,Angular displacementrad,MoloToAngular frequency (angular displacement/time)rads–1T–1Angular impulse (torque x time)NmsML2T–1Angular momentum (Iω)kgm2s–1ML2T–1Angular velocity (angle/time)rads–1T–1Area (length x breadth)m2L2Boltzmann’s constantJK–1ML2T–2θ–1Bulk modulus (\Delta P.\frac{V}{\Delta V}ΔP.ΔVV​.)Nm–2, PaM1L–1T–2Calorific valueJkg–1L2T–2Coefficient of linear or areal or volume expansionoC–1 or K–1θ–1Coefficient of surface tension (force/length)Nm–1 or Jm–2MT–2Coefficient of thermal conductivityWm–1K–1MLT–3θ–1Coefficient of viscosity (F =\eta A\frac{dv}{dx}ηAdxdv​)poiseML–1T–1Compressibility (1/bulk modulus)Pa–1, m2N–2M–1LT2Density (mass / volume)kgm–3ML–3Displacement, wavelength, focal lengthmLElectric capacitance (charge/potential)CV–1, faradM–1L–2T4I2Electric conductance (1/resistance)Ohm–1 or mho or siemenM–1L–2T3I2Electric conductivity (1/resistivity)siemen/metre or Sm–1M–1L–3T3I2Electric charge or quantity of electric charge (current x time)coulombITElectric currentampereIElectric dipole moment (charge x distance)CmLTIElectric field strength or Intensity of electric field (force/charge)NC–1, Vm–1MLT–3I–1Electric resistance (\frac{potential\text{ difference}}{current}currentpotential difference​)ohmML2T–3I–2Emf (or) electric potential (work/charge)voltML2T–3I–1Energy (capacity to do work)jouleML2T–2Energy density (\frac{energy}{volume}volumeenergy​)Jm–3ML–1T–2Entropy (\Delta S=\Delta Q/TΔS=ΔQ/T)Jθ–1ML2T–2θ–1Force (mass x acceleration)newton (N)MLT–2Force constant or spring constant (force/extension)Nm–1MT–2Frequency (1/period)HzT–1Gravitational potential (work/mass)Jkg–1L2T–2Heat (energy)J or calorieML2T–2Illumination (Illuminance)lux (lumen/metre2)MT–3Impulse (force x time)Ns or kgms–1MLT–1Inductance (L) (energy =\frac{1}{2}L{{I}^{2}}21​LI2) or coefficient of self-induction henry (H)ML2T–2I–2Intensity of gravitational field (F/m)Nkg–1L1T–2Intensity of magnetization (I)Am–1L–1IJoule’s constant or mechanical equivalent of heatJcal–1MoLoToLatent heat (Q = mL)Jkg–1MoL2T–2Linear density (mass per unit length)kgm–1ML–1Luminous fluxlumen or (Js–1)ML2T–3Magnetic dipole momentAm2L2IMagnetic flux (magnetic induction x area)weber (Wb)ML2T–2I–1Magnetic induction (F = Bil)NI–1m–1 or TMT–2I–1Magnetic pole strength (unit: ampere–meter)AmLIModulus of elasticity (stress/strain)Nm–2, PaML–1T–2Moment of inertia (mass x radius2)kgm2ML2Momentum (mass x velocity)kgms–1MLT–1Permeability of free space (\mu_o = \frac{4\pi Fd^{2}}{m_1m_2}μo​=m1​m2​4πFd2​)Hm–1 or NA–2MLT–2I–2Permittivity of free space ({{\varepsilon }_{o}}=\frac{{{Q}_{1}}{{Q}_{2}}}{4\pi F{{d}^{2}}}εo​=4πFd2Q1​Q2​​.)Fm–1 or C2N–1m–2M–1L–3T4I2Planck’s constant (energy/frequency)JsML2T–1Poisson’s ratio (lateral strain/longitudinal strain)––MoLoToPower (work/time)Js–1 or watt (W)ML2T–3Pressure (for

ammonium cyanate

Urea

No question bank is right to read you have to just see your headings carefully. And just remind about your previous learning about that topic. It is one of the best way to review before examination. I hope if you will do it, you will get better result.

No , just take a look to all topic , and read the topic you didn't prepared yet. And yes, don't forget to rewind each and Every formula ,,
☻☻To bas padhai karte rahe, All the very best ☻☻

through this periodic table Group12 3456789101112131415161718Period 1 1 H 1.008 Hydrogen 2 He 4.0026 Helium 2 3 Li 6.94 Lithium 4 Be 9.0122 Beryllium 5 B 10.81 Boron 6 C 12.011 Carbon 7 N 14.007 Nitrogen 8 O 15.999 Oxygen 9 F 18.998 Fluorine 10 Ne 20.180 Neon 3 11 Na 22.990 Sodium 12 Mg 24.305 Magnesium 13 Al 26.982 Aluminium 14 Si 28.085 Silicon 15 P 30.974 Phosphorus 16 S 32.06 Sulfur 17 Cl 35.45 Chlorine 18 Ar 39.948 Argon 4 19 K 39.098 Potassium 20 Ca 40.078 Calcium 21 Sc 44.956 Scandium 22 Ti 47.867 Titanium 23 V 50.942 Vanadium 24 Cr 51.996 Chromium 25 Mn 54.938 Manganese 26 Fe 55.845 Iron 27 Co 58.933 Cobalt 28 Ni 58.693 Nickel 29 Cu 63.546 Copper 30 Zn 65.38 Zinc 31 Ga 69.723 Gallium 32 Ge 72.630 Germanium 33 As 74.922 Arsenic 34 Se 78.971 Selenium 35 Br 79.904 Bromine 36 Kr 83.798 Krypton 5 37 Rb 85.468 Rubidium 38 Sr 87.62 Strontium 39 Y 88.906 Yttrium 40 Zr 91.224 Zirconium 41 Nb 92.906 Niobium 42 Mo 95.95 Molybdenum 43 Tc ☢ 96.906 Technetium 44 Ru 101.07 Ruthenium 45 Rh 102.91 Rhodium 46 Pd 106.42 Palladium 47 Ag 107.87 Silver 48 Cd 112.41 Cadmium 49 In 114.82 Indium 50 Sn 118.71 Tin 51 Sb 121.76 Antimony 52 Te 127.60 Tellurium 53 I 126.90 Iodine 54 Xe 131.29 Xenon 6 55 Cs 132.91 Caesium 56 Ba 137.33 Barium * 71 Lu 174.97 Lutetium 72 Hf 178.49 Hafnium 73 Ta 180.95 Tantalum 74 W 183.84 Tungsten 75 Re 186.21 Rhenium 76 Os 190.23 Osmium 77 Ir 192.22 Iridium 78 Pt 195.08 Platinum 79 Au 196.97 Gold 80 Hg 200.59 Mercury 81 Tl 204.38 Thallium 82 Pb 207.2 Lead 83 Bi 208.98 Bismuth 84 Po ☢ 208.98 Polonium 85 At ☢ 209.99 Astatine 86 Rn ☢ 222.02 Radon 7 87 Fr ☢ 223.02 Francium 88 Ra ☢ 226.03 Radium ** 103 Lr ☢ 262.11 Lawrencium 104 Rf ☢ 267.12 Rutherfordium 105 Db ☢ 270.13 Dubnium 106 Sg ☢ 269.13 Seaborgium 107 Bh ☢ 270.13 Bohrium 108 Hs ☢ 269.13 Hassium 109 Mt ☢ 278.16 Meitnerium 110 Ds ☢ 281.17 Darmstadtium 111 Rg ☢ 281.17 Roentgenium 112 Cn ☢ 285.18 Copernicium 113 Nh ☢ 286.18 Nihonium 114 Fl ☢ 289.19 Flerovium 115 Mc ☢ 289.20 Moscovium 116 Lv ☢ 293.20 Livermorium 117 Ts ☢ 293.21 Tennessine 118 Og ☢ 294.21 Oganesson  *Lanthanoids* 57 La 138.91 Lanthanum 58 Ce 140.12 Cerium 59 Pr 140.91 Praseodymium 60 Nd 144.24 Neodymium 61 Pm ☢ 144.91 Promethium 62 Sm 150.36 Samarium 63 Eu 151.96 Europium 64 Gd 157.25 Gadolinium 65 Tb 158.93 Terbium 66 Dy 162.50 Dysprosium 67 Ho 164.93 Holmium 68 Er 167.26 Erbium 69 Tm 168.93 Thulium 70 Yb 173.05 Ytterbium **Actinoids

He examined the relationship between atomic weights of the elements and their physical and chemical properties.Among chemical properties, Mendeleev mainly concentrated on the compounds formed by the elements with hydrogen and oxygen because they are highly reactive and hence formed compounds with almost all the elements.

The formulae of the hydrides and oxides formed by the various elements was made the basis of classification of elements.

Mendeleev in 1869, proposed periodic law.

It states that

The physical and chemical properties of elements are a periodic function of their atomic weights i.e. when the elements are arranged in order of their increasing atomic weight, elements with similar properties are repeated after certain regular intervals. This repetition of properties of elements after certain regular intervals is called periodicity of properties.

Characteristics of Mendeleev’s periodic table

1)Mendeleev’s arranged the then known elements in order of their increasing atomic weights, grouping together elements with similar properties and leaving out blank space wherever necessary.

2)He also made the prediction that there were some unknown elements which, would be discovered in due course of time and would fill these blank spaces.

3)He predicted their properties in the light of the properties of the other elements in the same group. Later on, these unknown elements were discovered and were found to possess exactly the same properties as predicted by Mendeleev.

4)Noble gases were not known at the time of Mendeleev. When these gases were discovered, a new group called the zero group was added toMendeleev original periodic table.

5)Two group of 14 elements each, called lanthanides and actinides were placed at the bottom of the periodic table.

6)Mendeleev’s periodic table consists of :

periodsa) 9 vertical columns called groups. These are designated as 0, , , , ,,  , , 

b) Except for group 0 and  ,each group is further divided into two subgroup designated as A and B.The elements which lie on the left hand side of each group constitutes sub-group A while those placed on the right hand side from sub-group B.This subdivision is made on the basis of the differences in their properties.

c)Group contains 9 elements in three sets each containing three elements. Group 0 has no subgroup. It consists of only one vertical column of inert gases.

d)7 horizontal rows called periods.These are numbered from 1 to 7.

Very short answer .... Discharge tube experiment led to the discovery of electron William crooks performed the discharge tube experiment by using two metallic rodes.When the discharge tube filled with the inert gas like helium at very low pressure and high voltage is applied some new types of negatively charged particles rays emitted from the cathode called as cathode rays.And the negativily charged particles are known as electron.. I am not just doing copy paste its my own language

DISCOVERY OF ELECTRON:

The Electron was discovered by J.J Thomson by conducting a Cathode ray tube experiment.

For the experiment he used Crooke’s tube, which was 60cm long glass tube and had a small tube attached. To this small tube vacuum pump was attached, it also had two metal plates which were connected to battery by wires.

The tube contained gas at atmospheric pressure. when current at high voltage (10,000volts) was passed following observations were made:

1. When current was passed through a gas at 1 atmospheric pressure and at a very high voltage, nothing happened. That is no visible effect was seen inside the tube.
2. Then further the pressure of gas was reduced by pumping the air out, with the help of vacuum pump. The pressure was reduced to 10^-2atm, then on passing current it was seen that whole tube started glowing green.
   Then further the pressure was reduced to 10^-4It was seen the whole glow vanished, but it was seen that at the end of the tube (anode side) there was a faint green glow observed.

To confirm the faint glow anode was made perforated, and a zinc sulphide screen (fluorescent material) was placed behind it.

When current was passed under same conditions it also started   glowing green. This confirmed that under those conditions some rays were emitted through cathode, and were travelling towards anode. Those rays were called as cathode rays and found to consist of negatively charged particles called electron.

Since the line in Hydrogen spectrum lies within region that is at Red end.

 Therefore it is corresponds to the Balmer series.
 The line at the Red end suggests that :-

 The 1st line of Balmer series is => n=3 to n=2

 The 2nd line of Balmer series is => n=4 to n=2

 The 3rd line of Balmer series is => n=5 to n=2
 Thus the answer is => 5 --> 2 = 5 to 2

Thank you

Transition is from n=5 to n=2

$$Be^+$$ while $$Be^{+3}$$ is applicable, an atom should be hydrogen like to be applicable.....

Be+

For brachett series n1 =4 and n2=5,6.... Infinite

Wrong , it jumps to 4 energy level, from 5 to $$\infty$$

N1=4 it jumps from 4 energy level

Chemistry is a branch of science which studies the structure of substance atoms molecules ions their composition and formation

There are two ways of balancing Redox reactions:

• Oxidation number method
• Half equation method

Oxidation method: The steps to be followed-

1. Write the skeletal equation of reactants and products.
2. Indicate the oxidation number of all the elements involved in the reaction.
3. Calculate the increase or decrease in oxidation number per atom. Also, identify the oxidizing and reducing agents.

1. Multiply the formula of oxidizing agent and reducing agent by suitable integers, so as to equalize the total increase or decrease in oxidation number as calculated in step c.
2. Balance all atoms other than H and O.
3. Finally balance H and O atoms by adding water molecules using hit and trial method.
4. In case of Ionic reactions:

• For acidic medium
• First balance O atoms by adding water molecules to the deficient side.
• Balance H+ ions to the side deficient in H atoms.
• For Basic medium
• First balance oxygen atom by adding water molecules to the deficient side.
• Then to balance hydrogen, add water molecules equal to the number of deficiency of H atoms.
• Also add equal number of OH^- ions to opposite side of the equation.

Example: Permagnate ion reacts with bromide ion in basic medium to give manganese dioxide and Bromate ion .

Step1: the skeletal ionic equation is :

MnO₄^- (aq) +Br^- (aq) ---> MnO₂ +BrO₃^-

Step 2: assign oxidation numbers for Mn and Br

Step3: calculate the increase and decrease in oxidation number and make the change equal :

Step: 4 as the reaction occurs in basic medium, and the ionic charges are not equal on both sides, add 2OH^- ions on the right to make it equal.

Step5: finally count the hydrogen atoms and add appropriate number of water molecules on the left side to achieve balanced Redox reaction.

Half reaction method or Ion electron method

1. Write the skeletal equation and indicate the oxidation number of all the elements which occur in skeletal equation
2. Find out the species that are oxidized and reduced.
3. Split the skeletal equation into two half reactions: oxidation half reaction and reduction half reaction
4. Balance the two-half equation separately by rules described below:

• In each half reaction first balance the atoms of element that has undergone a change in oxidation number.
• Add electrons to whatever side is necessary to make up the difference in oxidation number in each half reaction.
• Balance the charge by adding H⁺ ions, if the reaction occurs in acidic medium .For basic medium, add OH^- ions if the reaction occurs in basic medium.
• Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms
• In the acidic medium, H atoms are balanced by adding H ⁺ ions to the side deficient in H atoms.
• However, in the basic medium H atoms are balanced by adding water molecules equal to number to H atoms deficient.
• Add equal number of OH^- ions to opposite side of equation.
• The two half reactions are then multiplied by suitable integers .so that the total number of electrons gained in half reaction becomes equal to total number of electrons lost in another half reaction.
• Then the two half reactions are added up.
• To verify the balancing, check whether the total charge on either is equal or not.

Example: Let us consider the skeletal equation:

Fe²⁺ + Cr₂O₇^2- -->  Fe³⁺ +Cr³⁺

Step 1: Separate the equation in to two halves:

Oxidation half reaction: Fe² --> Fe³⁺

 Reduction half reaction: Cr₂O₇^2-  -->    Cr³⁺

Step 2: Balance the atoms other than hydrogen and oxygen in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms .For the reduction half reaction, we multiply the Cr³⁺ by 2 to balance Cr atoms.

Step 3: For reactions occurring in acidic medium, add water molecules to balance oxygen atoms and hydrogen ions are balanced by adding H atoms. Thus, we get:

Cr₂O₇^2- + 14 H⁺  +  6e^- --> 2 Cr³⁺ + 7H₂O

Step 4: Add electrons to one side of the half reaction to balance the charges .if needed make the number of electrons equal in two half reactions by multiplying one or both half reaction by suitable coefficient.

The oxidation half reaction is thus written again to balance the charge .Now in the reduction half reaction there are 12 positive charges on the left hand side and only 6 positive charge on right hand side .Therefore, we add six electrons to left hand side .

Cr₂O₇^2- + 14 H⁺  +  6e^- --> 2 Cr³⁺ + 7H₂O

To equalize the number of electrons in both reactions, we multiply oxidation half reaction by 6 and write as:

6Fe²⁺  -->  6Fe³⁺ +6e^-

Step 5: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side .This give us net ionic equation:

6Fe²⁺  + Cr₂O₇^2- + 14 H_⁺ -->    2Cr³⁺+6Fe³⁺ +7H₂O

Step6: Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number atoms and the charges.

3 period 17 group

Chlorine has an atomic number 17 in the periodic table

$\underset{x_i}{\overset{x_f}{\oint }}\overrightarrow{F}.d\overrightarrow{x}$

Since lone pairs are placed at equitorial position so this structure is unstable because according to the bent's rule the electronegative elements must placed at equitorial position in compound with Trigonal bypriamidal geometry not the lone pairs.

???

Hahajaja?

Thanks mam and you are elder so plz dont be sorry

It can be defined as the electrons which are there in degenerate orbitals have a parallel spin and tend to exchange their position.  The exchange energy is the energy released when two or more electrons with the same spin-exchange their positions in the degenerate orbitals of a subshell. Exchange energy is nothing but the energy released during this process. When the orbitals are half-filled or completely filled then the number of exchanges is maximum. Therefore, its stability is maximum.

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$$

−b±b2−4ac−−−−−−−√2a $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The wave function (psi)for a moving object is not a observable quantity but psi square has a physical significance which tells about the area in which the probability of getting electrons is 90 percent

Ionisation enthalpy: It is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom.
M(g) →M⁺(g) + e^–I₁  = First ionisation enthalpy
Similarly, second and third electrons are also removed by providing successive ionisation enthalpies.
Factors on which Ionisation Enthalpy Depends:
(i) Size of the atom: The larger the atomic size, smaller is the value of ionisation enthalpy. In a larger atom, the outer electrons are far away from the nucleus and thus force of attraction with which they are attracted by the nucleus is less and hence can be easily removed.
Ionisation enthalpy ∝ 1/ Atomic size
(ii) Screening effect: Higher the screening effect, the lesser is the value of ionisation enthalpy as the screening effect reduces the force of attraction towards nucleus and hence the outer electrons can be easily removed.
Ionisation enthalpy ∝  1/ Screening effect
(iii) Nuclear charge: As the nuclear charge increases among atoms having same number of energy shells, the ionisation enthalpy increases because the force of attraction towards nucleus increases.
Ionisation enthalpy ∝ Nuclear charge
(iv) Half filled and fully filled orbitals: The atoms having half filled and fully filled orbitals are comparatively more stable, hence more energy is required to remove the electron from such atoms. The ionization enthalpy is rather higher than the expected value in case of such an atom.
Ionisation enthalpy ∝ Stable electronic configuration
(v) Shape of orbital: The s-orbital is more close to nucleus than the p-orbital of the same orbit. Thus, it is easier to remove electron from a p-orbital in comparison to s-orbital. In general, the ionisation enthalpy follows the following order
(s>p> d>f) orbitals of the same orbit.
Variation of ionisation enthalpy in the periodic table
ln general, the ionisation energy decreases down the group due to increase in atomic size. On the other hand, the ionisation energy increases across the period from left to right, again due to decrease in atomic size from left to right.

The energy of one photon is

E= hc/λ

 = (6.626×10⁻³⁴Js×3×10⁸m/s​)/ 0.57×10⁻⁶

 =3.487×10⁻¹⁹J .

25 watt corresponds to 25 J per second.

The number of photons emitted per second is 25​ / 3.487×10⁻¹⁹=7.17×10¹⁹/s. 

Air contains 20% oxygen by volume. Calculate the theoretical volume of air which will be required for burning completely 500 m³ of acetylene gas. All volumes are measured under the same conditions of temperature and pressure.

A n s w e r

(d) pentagonal bipyramid ?

 Iodine heptafluoride, also known as iodine(VII) fluoride or iodine fluoride, is an interhalogen compound with the chemical formula IF7. It has an unusual pentagonal bipyramidal structure, as predicted by VSEPR theory.

Balmer series

ANSWER
In Lyman series, the series of lines are in the Ultraviolet region 

In Balmer series, the series of lines are in the Visible region 

The series of lines in Paschen and Brackett series lies in an Infrared region

Hence, the correct option is Ad

According to law of conservation of mass,

Mass of reactants = Mass of products.

AgNO3 + NaCl = AgCl + NaNO3

xg 5.85g = 14.35g 8.5g

x 5.85 = 14.35 8.5

x. = 17.0g a n s w e r

According to law of conservation of mass,

Mass of reactants = Mass of products.

AgNO3 + NaCl = AgCl + NaNO3

xg 5.85g = 14.35g 8.5g

x 5.85 = 14.35 8.5

x. = 17.0g answer

First of all it was studied by Einstein that all microscopic particles shows dual nature i.e wave nature and particle nature .Then De Broglie co related these nature by an equation called as De Broglie equation