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1. Recognise the functional group in the compound. This will determine the suffix of the name 

   Functional group	suffix
   alkane	-ane
   alkene	-ene
   alkyne	-yne
   alcohol	-ol
   aldehyde	-al
   ketone	-one
   carboxylic acid	-oic acid
   ester	-oate

    The suffix associated with various functional groups.

2. Find the longest continuous carbon chain that contains the functional group (it won't always be a straight chain) and count the number of carbon atoms in this chain. This number will determine the prefix (the beginning) of the compound's name .

   Carbon atoms	prefix
   1	meth-
   2	eth-
   3	prop-
   4	but-
   5	pent-
   6	hex-
   7	hept-
   8	oct-
   9	non-
   10	dec-

     The prefix of a compound's name is determined by the number of carbon atoms in the longest chain that contains the functional group.

3. Number the carbons in the longest carbon chain (Important: If the molecule is not an alkane (i.e. has a functional group) you need to start numbering so that the functional group is on the carbon with the lowest possible number). Start with the carbon at the end closest to the functional group.

4.  

   Look for any branched groups:

   • Name them by counting the number of carbon atoms in the branched group and referring totable he group on the main carbon chain. If there is more than one of the same type of branched group then both numbers must be listed (e.g. 2,4 -) and one of the prefixes listed in must be used. Important: If the molecule is an alkane the branched group must be on the carbon with the lowest possible number.
   • The branched groups must be listed before the name of the main chain in alphabetical order (ignoring di/tri/tetra).

   If there are no branched groups this step can be ignored.

   Number	prefix
   2	di-
   3	tri-
   4	tetra-

   Table 4.7: Prefixes for multiple substituents with the same name. These apply to multiple functional groups as well.

5. For the alkyl halides the halogen atom is treated in much the same way as branched groups:

   • To name them take the name of the halogen atom (e.g. iodine) and replace the -ine with -o (e.g. iodo).

     Halogen	name
     fluorine	fluoro
     chlorine	chloro
     bromine	bromo
     iodine	iodo

     Table 4.8: Naming halogen atoms in organic molecules.

   • Give the halogen atom a number to show its position on the carbon chain. If there is more than one halogen atom the numbers should be listed and a prefix should be used (e.g. 3,4-diiodo- or 1,2,2-trichloro-). See <a href="https://intl.siyavula.com/read/science/grade-12/organic-molecules/04-organic-molecules-03#tab:organic:prefix2">Table 4.7</a> for a list of the prefixes.
   • The halogen atoms must be listed before the name of the main chain in alphabetical order (ignore di/tri/tetra).

   If there are no halogen atoms this step can be ignored.

6. Combine the elements of the name into a single word in the following order:

   • branched groups/halogen atoms in alphabetical order (ignoring prefixes)
   • prefix of main chain
   • name ending according to the functional group and its position on the longest carbon chain.

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118

118

118

118 + 1 (eka-francium) = 119

118

Yess

• ISOTOPES: Are those elements which have same atomic number ,but different mass number.
• 
•  ISOBARS: Are those elements which have same mass number, but different atomic number.
• 

Having different atomic mass and same atomic no. And vice versa

Candela cd

he SI unit of luminous intensity is Candela and temperature is Kelvin.

Where is the question

Inductive effect is an effect in which permanent polarization arises due to partial isplacement of sigma e- along carbon chain or partial displacement of sigma-bonded electron toward more electronegative atom in carbon chain i.e. Magnitude of partial positive charge: Inductive effect is a permanent effect.

 Aromatic heterocyclic compounds: Aromatic cyclic compounds containing one or more heteroatoms in their molecules are called aromatic heterocyclic compounds. For example,

Sigma Bond	Pi Bond
During sigma bond formation overlapping orbitals can either be one hybrid orbital and a single pure orbital, or two pure orbitals and two hybrid orbitals.	During pi bond formation overlapping orbitals must be two unhybridized orbitals.
Sigma bonds are known to exist independently and allow free rotation.	Pi-bonds must always exist along with sigma bond and the rotation is restricted.
Sigma bonds are stronger bonds.	Pi bonds are usually less powerful compared to sigma bonds.
Sigma bonds are formed first when atoms interact.	Pi bonds between two atoms are formed after sigma bonds are formed between them.
During the bonding between two given atoms,   Only one sigma bond is formed.	Here two pi bonds can exist between two atoms.
Sigma bonds are known to have cylindrical charge symmetry around the axis of the bond.	No symmetry exists in pi bonds.

0.5 = mass/ molar mass 0.5 × molar mass = mass 0.5 × 44= mass Mass = 22 g

It is quite simple. U know well No of mole is = given weight divided by molecular weight U hv to calculate mass.and u know mole well and molecular mass of CO2 which is 44g/mol So mass becomes 22g

A) Work function of caesium (WO) = hv_o

Therefore v_o = W_O/h = 1.9 x1.602 x10^-19 / 6.626x10^-34

= 4.59x10¹⁴/sec

B) λ_o =c/v_o = 3x10⁸ / 4.59x10¹⁴ = 6.54x10^-7m

C) K.E of ejected electron = h(v-v_o) = hc(1/λ – 1/ λ_o)

=(6.626x3x10^-26) (1/500x10^-9 – 1/654x10^-9)

=(6.626x3x10^-26) / 10^-9(154/500x654)

= 9.36x10^-20J

k.E = 1mv²/2 = 9.36x10^-20J

=9.1x10^-31/2 = 9.36x10^-20J

Or

v2 = 20.55x10¹⁰m²s^-2

Or

v = 4.53x10⁵ms^-1

The Ostwald’s Dilution Law is defined for a weak electrolyte as “ the degree of ionization is proportional to the square root of the molar concentration or directly proportional to square root of the volume of the solution which contains one mole of the electrolyte.”

Mathematically, we can write Ostwald’s Dilution Law as below:e

                                           α = √ k_a/C   = √ k_aV

                                     Or, α = √ k_b/C   = √ k_bV

Wilhelm Ostwald’s dilution law is a relationship proposed in 1888 between the dissociation constant Kd and the degree of dissociation α of a weak electrolyte. The law takes the form Where the square brackets denote concentration, and c₀ is the total concentration of electrolyte. Ostwald's dilution law describes the dissociation constant of the weak electrolyte with the degree of dissociation (α) and the concentration of the weak electrolyte.  α = degree of dissociation. Ostwald's dilution law states that only at infinite dilution the weak electrolyte undergoes complete ionization.

• The boundary surface diagram for the s orbital looks like a sphere having the nucleus as its centre which in two dimensions can be seen as a circle.
• Hence, we can say that s-orbitals are spherically symmetric having the probability of finding the electron at a given distance equal in all the directions.
• The size of the s orbital is also found to increase with the increase in the value of the principal quantum number (n), thus, 4s > 3s> 2s > 1s.

3-2.9

According to me the correct answer would be 2.976

4d :
N = 4 , l = 2,

Because half filled shells are more stable than another one

4s² $$\boxed{↑\↓}$$ 3d⁴ $$\boxed{↑\ \|\↑\ \|\ ↑\ \|\↑\ \| \ \}$$
hear 4s is full filled so it is stable but 3d is not stable
<hr> 4s¹ $$\boxed{↑\ }$$ 3d⁵ $$\boxed{↑\ \|\↑\ \|\ ↑\ \|\↑\ \|↑\ \}$$
hear both 4s & 3d are more stable <hr><hr>

wait, are you sure you areDeepak Singh ?

Yeah see the energy order of subshells

Bcoz it's more stable for them to have half-filled D-subshells ?

Ionisation enthalpy is also known as ionisation potential since it is the minimum potential difference required to remove the most loosely bound electrons from an isolated gaseous cation.The energies required to knock out second and third electrons are called second and third ionisation energies.

अपने ऑर्बिट से electron ko hatane me kitna energy chahiye for infinite orbit tak then, uus energy ko ionisation potential karate hai

The amount of energy required to remove most loosely bounded electrons

Don't know

The inert pair effect is the tendency of the two electrons in the outermost atomic s-orbital to remain unshared in compounds of post-transition metals. The inert pair effect explains why common ions of Pb are Pb4+ and Pb2+, and not just Pb4+ as we might expect from the octet rule. Screening effect : The shielding effect describes the balance between the pull of electrons on valence electrons and the repulsion forces from inner electrons.

Happy Diwali

Happy deepawali

Ok happy diwali if I am friend?

It is a Columbic Force of attraction between oppositly charged ion which are formed by loss and gain electron

???

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules" ▤=▥=▦=▧=▨=▩

Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure. As most of the reactions are carried out at constant pressure ,the measured value of the heat evolved or absorbed is the enthalpy change enthalpy.

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An atomic unit of mass is defined as accurately 1/12 the mass of a carbon-12 atom. The carbon-12 atom has six neutrons and six protons in its nucleus. The atomic unit mass is symbolized as AMU or amu.

1 AMU = Average of the proton rest mass and the neutron rest mass.

1 AMU = 1.67377 x 10 ^-27 kilogram or 1.67377 x 10 ^-24 gram.

4s $$\boxed{↑\↓}$$ 3d $$\{↑\ \|\↑\ \|\ ↑\ \|\↑\ \| \ \}$$ 4s $$\boxed{↑\↓}$$ 3d $$\↑\ \|\↑\ \|\ ↑\ \|\↑\ \| \ \$$

4s $$\boxed{↑\↓}$$ 3d $$\boxed{↑\ \|\↑\ \|\ ↑\ \|\↑\ \| \ \}$$ 4s $$\boxed{↑↓}$$ 3d $$\boxed{↑\ \|\↑\ \|\ ↑\ \|\↑\ \| \ \}$$

4s $$\boxed{↑\↓}$$ 3d $$\boxed{↑\ \|\↑\ \|\ ↑\ \|\↑\ \| \ \}$$ 4s $$\boxed{↑\↓}$$ 3d $$↑\ \|\↑\ \|\ ↑\ \|\↑\ \| \ \$$

(2) 1 inch = 3.33cm is the incorrect one. Actually 1 inch = 2.54 cm

Intermixing of different level of energy of orbital is known as HYBRIDIZATION

seeing you after a long time , where was you??

We know orbitals of last shell overlap with each other.  The overlapping is of two types:

• Head to head(sigma bond)
• Sidewise(pi bond)

Overlapping of orbitals takes place between which has same energy. If in case, the orbitals have different energy they can’t overlap. Hybridization is the intermixing of orbitals of slightly different energies, so as to redistribute their energy and give rise to new set of orbitals that are similar in shapes and energy.

Intwr mixing of orbitals to firm bonding orbitala

Resonance is the mixture of all Lewis dot structure of compound

How much of 0.3 M ammonium hydroxide should be mixed with 30 ml of 0.2 M solution of ammonium chloride to give buffer solutions 10 pH 8.65 and 10. Given pK_b of NH₄OH = 4.75.

A n s w e r :

Let V mL of NH₄OH be mixed with NH₄CI to have a buffer of pH 8.65. The total volume after mixing becomes (V+ 30) mL.

Let V mL of NH4OH be mixed with NH4CI to have a buffer of pH 8.65. The total volume after mixing becomes (V+ 30) mL.

m Mole of NH4OH = 0.3×V

Thus, NH4OH = 0.3×V/ (v+30)

m Mole of NH4CL = 0.2×30

Thus, NH4CL  = 0.2×30/ (v+30)

= 0.2×30/ (v+30)/ 0.3×V/ (v+30)

= 0.61 = log6 /0.3×V

= v = 4.91mL

Similarly,

14-10 = 4.74+log 0.2×30/ (v+30)/ 0.3×V/ (v+30)

for ph = 10

V = 109.9mL

Mai vdiya hu.. But padhai n likhai nhi hai ???aap btao??

Lewis dot structures reflect the electronic structures of the elements, including how the electrons are paired. Lewis structures are a useful way to summarize certain information about bonding and may be thought of as “electron bookkeeping”. In Lewis dot structures each dot represents an electron. A pair of dots between chemical symbols for atoms represents a bond.

Lewis Structure of CO (Carbon Monoxide)

• A carbon monoxide molecule consists of one carbon atom and one oxygen atom.
• The carbon atom requires four electrons to obtain octet configuration whereas the oxygen atom requires two.
• Therefore, the valency is satisfied via the donation of a lone pair of electrons for bonding by the oxygen atom.
• The resulting Lewis electron dot structure displays a triple bond connecting a carbon and an oxygen atom, each holding a lone pair of electrons.

 

Hi goog morning

According to Bohr,

• When energy is supplied to atoms of hydrogen, the electron from lower energy gets excited to higher energy level.
• The excited state being unstable, it jumps back to its original state that is ground state.
• Some electrons move to their ground state in one jump, some in multiple jumps. Each jump corresponds to line in a spectrum.
• As we know the gas in tube consists of many hydrogen atoms.
• Therefore, each electron on getting energy gets excited.
• On returning to the ground state, they either move in single jump or multiple jump.
• This is the reason that we get so many lines in different regions in hydrogen spectrum.

The wavelength emitted by them can be calculated as: