{tex}1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}\\5s^25p^64f^{14}5d^56s^2{/tex}

Partial equation method:

Partial equation method is used for balancing complex chemical equations. There are chemical equation which involves many steps and reacting elements occur in more than one products in the product side. In such cases, balancing by Hit and Trial method is difficult. That is why partial equation method is used to balance such chemical equations.

In Partial equation method, equation is first divided into partial equation, which are simply probable steps that might occur in the chemical reaction. These probable steps are then balanced by hit and trail method and finally added. Following steps are applied in balancing chemical equation by partial equation method.

1. At first, different probable steps are written for the given chemical equation. These probable steps are simply the partial chemical equation which we can write for easily balancing the reactants and product. Note that the probable steps may or may not occur in real.
2. After breaking down the probable steps for the chemical equation, the partial equations are individually balanced by using Hit and Trial Method.
3. Such balanced partial equations are multiplied by suitable integer. This is done if required, so that the elements which are not formed in the product side of the overall chemical equation is canceled out.
4. Finally, the balanced partial equations are added to get the final overall balanced equation.

Lets’ illustrate some examples:

PbS + O3    →PbSO4 + O2

In the given reaction, the atom of Pb and S is balanced in both reactant and product side. But there is 3 atom of oxygen in reactant side and 6 oxygen atom in product side. So let’s balance this chemical equation by partial equation method.

• First let’s think the possible step for occurring of the given chemical reaction. There is lead sulphide reacting with ozone, it means ozone must be acting as oxidizing agent which liberates nascent oxygen. So the first step can be generation of nascent oxygen by decomposition of ozone molecule.

Step-1

O3  →O2 + [O]

This liberated nascent oxygen can react with PbS to give lead sulphate and oxygen as;

Step-2

PbS + [O] → PbSO4

This partial step is not balanced reaction. There is equal number of Pb and S atom, but no. of oxygen is one in reactant side but four in product side. So we can multiply nascent oxygen in reactant side by four.

PbS + 4[O] → PbSO4

If we make four nascent oxygen in second probale step, then we must also need to make it equal number of nascent oxygen liberated in step one. So we need to multiply step one with integer four

{ O3  →O2 + [O]    }    × 4

Now lets add both partial equation:

{ O3   →O2 + [O]    }  × 4

PbS + 4[O] → PbSO4

PbS  + 4O3   PbSO4 + 4O2

This is the balanced chemical equation.

Hybridization happens when atomic orbitals mix to form new atomic orbitals. The new orbitals have the same total electron capacity as the old ones.  The properties and energies of the new, hybridized orbitals are an 'average' of the original unhybridized orbitals.

The distinction began to emerge when a clear differentiation was made between chemistry and alchemy by Robert Boyle in his work The Sceptical Chymist (1661)

Robert boyle is the father of chemistry.

There are three rules on determining how many significant figures are in a number: 

• Non-zero digits are always significant.
• Any zeros between two significant digits are significant.
• A final zero or trailing zeros in the decimal portion ONLY are significant.

So keeping these rules in mind we can say that 11 is significant figure.

Shielding effect can be defined as a reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons on the nucleus. It is also referred to as the screening effect (or) atomic shielding.

Molality vs. Molarity:

These two words sound similar but they are not synonyms, even though both of them are used for representing solution concentration. By definition, molarity is the number of moles of solute dissolved per liter of solution. We use capital letter “M” to represent molarity and its formula is M= (# mol SOLUTE)/ (Liters of SOLUTION).

Molality is then the number of moles of solute per kilogram of the SOLVENT, NOT solution! We use lower case letter “m” to represent molality and its formula can be represented as: m= (# mole SOLUTE) / (Kilograms of SOLVENT).

Most of the time scientists use either molarity or molality to represent solution concentration, but MOLALITY is preferred when the temperature of the solution varies. That is because MOLALITY does not depend on temperature, (Neither number of moles of solute nor mass of solvent will be affected by changes of temperature.) while MOLARITY changes as temperature changes. (Volume of solution in the formula changes as temperature changes, and that is why.)

https://youtu.be/lEjQKATnI2A

Chlorine has higher electron gain enthalpy than fluorine because due to small size of fluorine some energy is utilized in overcoming the force of repulsion among electrons. Hence it has lesser value of electron affinity.

Dioxygen means O_{​​​​​​2} Oxygen basically exists as a diatomic gas, that is two oxygen atoms chemically combine to form an oxygen molecule which is denoted by 

O_{​​​​​​2} In other words, dioxygen is the elemental form of oxygen. So, if in a question only oxygen gas has been mentioned then we will consider O_{​​​​​​2} and take 32g as the molar mass because atomic mass of oxygen = 16u

So molecular mass of of oxygen gas (O_{​​​​​​2}) = 2 X 16 = 32 u and hence molar mass of oxygen gas = 32g

Otherwise, if oxygen molecule is mentioned, then it is quite clear that you will consider dioxygen and take 32g as the molar mass. However, if in the question specifically oxygen atom has been mentioned, then O atoms will be considered and the molar mass will be taken as 16g, because here we are talking about oxygen atoms, and not oxygen molecules.

Increase

B) Increase

Explanation : When you add pure water to an acidic solution, the solution becomes less acidic and the pH goes up -- to a point. For example, if you add pure water to a relatively acidic solution that has a pH of 3.5, the pH level might go up to a 4 or 5. You cannot, though, turn an acidic solution into a base solution or make it neutral just by adding water. As a result, pure water can only raise the pH of an acidic solution to a maximum of a 6.9.

Avogadro’s assertion is a “law” rather than a “hypothesis” or a “theory” precisely because it has withstood two centuries of rigorous usage and observation. Indeed the point of the law is that the volume of a gas does NOT depend on the size of the molecules. That said, it has been recognized from the start that Avogadro’s Law presumes an “ideal gas”. Deviations from ideal behavior are expected, for example, when gas molecules reversibly interact with one another or when gases are highly compressed.

J.J Thomson gave the first model of an atom in which its model is compared with a water melon or Christmas pudding. The reddish part of water melon represents positive charge ( protons ) and seeds embedded in it reflect negative charge ( electrons) . The total amount of positive charge is equal to total amount of negative charge. An atom as a whole is electrically neutral.

Limitations:- 

1. This model didn't get experimental support and hence discarded.

2. This model could not explain the stability of an atom as there is no boundary of separation between positive and negative charge

It is Krypton. Atomic number is 36.

Electronic configuration:

{tex}1s^2 2s^22p^6 3s^23p^64s^23d^{10}4p^6{/tex}

 ​Number of electrons in s orbital are 8 Number of electrons in p orbital are 18 Number of electrons in d orbitals are 10 (18-8=10)

STP stands for Standard Temperature and Pressure.
NTP stands for Normal Temperature and Pressure.

At STP: Pressure = 1 bar = 0.987 atm
Temperature = 273 K or 0°C

At NTP: Pressure = 1 atm
Temperature = 293 K or 20°C

d) SO₂

b)Fructose

Rocks

Nylon

No of waves are 3 

{tex}(mV)_A={h\over λ_A }{/tex}

{tex}(mV)_B={h\over λ_B }{/tex}

{tex}(mV)_B={1\over 2}(mV)_A {/tex}   [Given]

{tex}{h\over λ_B}={1\over 2}\times {h\over λ_A }{/tex}

{tex}λ_B=2λ_A {/tex}

=2×5×10⁻⁸ 

=10×10⁻⁸m

=1000A^∘

the question written over is incorrect.

the actual molecule should be like this H₂C=CH-CH₃

in the case of above molecule the C₁=SP² hybridisation

C₂=SP² hybridisation

C₃=SP³ hyridisation