Solution: 

{tex}AuCl_3 + NaCl \to Na[AuCl_4]{/tex}

Silver chloride dissolves in methyl amine solution because it forms soluble complex.

{tex}2CH_3NH_2 + AgCl \to {/tex}

{tex}[CH_3 -H_2N \to Ag \leftarrow NH_2-CH_3]Cl{/tex}

step 1,

Molality (m)  =   No. of  moles / 1000 g of solvent ( water )

...................=  1.25 / 1000 g

.................. = ( 1.25 x40  ) g /  1000 g

.................. =  50 g /  1000 g 

{tex}\therefore{/tex} total weight of the solution =  1050 g

step 2 ,

Percent by weight of the solution  is given by the expression ,

......................................................................................( Mass of solute ) / [ ( mass of solute )  +  ( mass of solvent ) ]   

Since  , 1050 g of solution contains  solute ( NaOH )   =   50.00  g

{tex}\therefore{/tex}   100 g of solution would contain solutte ( ie. NaOH ) = ( 50 / 1050 )

.................................................................................= 0.0476

Thus  , the percent  by weight of NaOH in the solution  is  0.0476 %

Atomic number of potassium is 19.

AMORPHOUS SOLIDS :-

• It has irregular shape
• isotropic in nature
• has short range order
• they are pseudo solids or super-cooled liquids
• When cut with a specific sharp edge tool, they cut into two pieces with irregular surfaces 

But the question is this much only and is from vibrant academy

No ur method is not correct. Because electrons removed from outermost shell that is 4s in case of Fe.

But if we look like Fe²⁺  as Cr  and

Fe³⁺ as V

in that sense then Fe²⁺ should be the correct,,,,

is my method correct?????

Electronic Configuration of

Fe = {tex}1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2{/tex}

Electronic Configuration of

Fe²⁺  = {tex}1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 {/tex}

Electronic Configuration of

Fe³⁺ = {tex}1s^2 2s^2 2p^6 3s^2 3p^6 3d^5{/tex}

So Fe³⁺ has more unpaired electron. 

A simple advice , ( of course  not a trick )

One shouldn't be in search  of  short cuts or tricks to understand or develop confidence over  descriptive topics like  " d - & f-block  elements" . The properties or charecteristics of d -  &  f -  block elements are  very well understood and explained using a relation between their electronic configurations and application of involved screening effects .Of course, there happens  to be exceptions in a few cases only ,which are needed to be kept in mind , eventhough these have their own justifications. 

 Conversion of a galvanic cell into an electrolytic cell is practically  possible  by  -

(i) connecting the cell to an external source of electricity , and

(ii) providing enough  potential difference through this external source  ,   such that  ,  the applied emf  exceeds  the  emf  of  the galvanic cell  itself. [ for example  >  1.1 V   ,in case of   simple Daniel cell  ie.(  Zn - Cu ) -cell. ]

Explanation :

Thus , In the   beginning the external opposing potential is less than 1.1 V.   In this case the normal cell reaction  occurs in the galvanic cell  and the current flows from copper electrode to zinc electrode. But when the opposing potential is  increased slowly and gradually it is observed that normal reaction continues to occur till the opposing potential reaches  to value of 1.1 V. At this stage the cell reaction stops .& the current continues to flow . Further increase in the opposing potential ( beyond 1.1 V ) again causes the chemical reaction to occur. But now  the reaction occurs in opposite direction forcing  Zn²⁺  ions to get reduced and Cu atoms to get oxidised.

Zn²⁺ (aq)   +     Cu (s)  ---------->     Zn (s)     +      Cu²⁺ (aq )

A galvanic cell is an electrochemical cell that uses the transfer of electrons in  redox reaction  to supply an electric current.This cell is driven  by spontaneous chemical  reaction that produces an electric current through an  outside  electrical circuit .Galvanic cell reactions supply energy  , which is used to perform work. For this reason galvanic cells are used  as  batteries .

 Galvanic cells harness the electrical  energy available  from the electron transfer  in a redox reaction to perform useful electrical work. The key to gathering the electron flow is  to separate the oxidation and reduction  half reactions., connecting them by a wire , so that the electrons must flow through that wire.  The electron flow called a current   can be sent through a circuit which could be part of  any number of electrical devices  such as radios ,  televisions , watches  etc.

A schemetic line diagram of  a galvanic cell may be represented in a simple way as-

............. Oxidation half cell --------> Salt bridge <---------Reduction half cell

                             |____________connecting wire_____________|

The salt bridge :   The salt bridge or a porous disk is necessary to maintain the charge neutrality of each half cell by allowing the flow of ions with minimal mixing of the half cell solutions .  As electrons are transferred from the oxidation half cell to the reduction half cell , a negative charge  builds

in the reduction half cell  and a positive charge in the oxidation half cell. The charge building up would serve to oppose the current from anode to cathode - effectively stopping the electron flow  - if the cell lacked a path for ions to flow between the two solutions. From  the schemetic diagram as illustrated above  it can easily be gathered that  that the electrode in oxidation half cell is called anode and the electrode in reduction half cell is called the cathode.. A good mnemonic to help remember that is " the Red  Cat  ate an Ox " meaning that reduction takes place at the cathode  and oxidation takes place at the anode .

The anode , as the source of negatively charged electrons is usually marked with a minus ( - ) sign and cathode is marked with a plus  ( + ) sign.

Line notation for a Galvanic Cell 

Instead of drawing a cell diagram  a shorthand way of describing  a cell is called  line notation. This  notation scheme places the constituents of cathode on the right and the anode components on the left. The phases   of all reactive species  are listed and their concentrations or pressures  are given if those species are not at standard states  ( ie. 1 atm . for gases  and 1M for solutions ). All phase  interfaces are noted with a single line (  |   ) and multiple species  in a single phase  are separated  by  commas. For example  a half cell containing  1M solution of CuO and  HCl and  a  Pt electrode  for reduction of Cu²⁺ would be written as :

Pt (s) |  Cu²⁺ ( aq ) , H⁺  ( aq )

Note that the spectator ions , oxide and chloride have been left out of notation  and that banode is written  to the far left.

The salt bridge or porous disk is shown in the notation  as  a double line  ( ||  ) . Therefore a cell that undergoes oxidation  of magnesium by  Al³⁺  would have the following cell notation if the anode is  magnesium  and the cathode is aluminium

Mg (s)  |  Mg²⁺ ( aq ) | |  Al³⁺ ( aq ) |  Al   (s)

Standard Reduction Potentials

One can measure the cell potential E_cell   , in volts , of any  galvanic cell using  a potentiometer. However it is directly impossible  to measure the potential of each individual half cell.. Hence , a method has been devised to measure thhe ability of a chemical species to reduce  by compiling  tables of standard reduction potentials E^o . Arbitrarily assigning a value of exactly zero to potential of  of the standard hydrogen electrode allows us tommeasure the E^o of any half reaction. That measurement is made by constructing  a galvanic cell between  the STandard Hydrogen Electrode (SHE) and the unknown half cell at  standard state conditions .For example  , if the following cell is constructed , an E^o _cell   of 0.34 V is observed

Pt (s ) |  H₂ (g) |  H⁺ ( aq )  | |  Cu²⁺  (aq)  |  Cu (s)

Because SHE has a potential of exactly zero volts , the reaction  Cu²⁺  + 2e  ------------>   Cu  has a value of  0.34 V for its  E^o  .  Note  that ,

E^o _cell  =  E^o SHE    +  E^o 

Fortunately , every important reduction potential has been measured  and tabulated . Useful list of reduction potentials are available in most of the introductory chemistry   texts , including yours. Using this list the EMF of galvanic cells are calculated.The following expression is helpful to calculate emf of a galvanic cell under Standard conditions  ( ie. 1 atm pressire  and  at  278 K ) 

E^o _cell    =    E^o _R   -   E^o _L  where  , R  &  L  mean  the  right  and  left   half cells when the line notation for  cell composition is written , repectively.

Amorphous solids don't break at fixed cleavage planes whereas Crystalline solids cleavage along particular direction at fixed cleavage planes.

Kernels are basically the atom without its valence shell .all the inner shells and the nucleus make up the kernel. The valence shell is represented outside the kernel.

This is example of Wurtz Reaction . Product ll be Bicyclobutane.

The standard enthalpy of formation for an element in its state is zero.

Solution

 The  posted question doesn't  mention  its nature or type . However , supposing that the question is an MCQ type question(  , the  option (ii)  is correct  ie.

' energy released during spontaneoous chemical reactions and the use of electrical energy to bring about nonspontaneous chemical transformations '

In LiCl structure the chloride ion forms a face centred  unit cell( anion - anion contact ). Hence chloride ions contact each other across the diagonal of the face. If we draw  a diagonal on the face of a unit cell , we find that -

length of the  diameter  = radii  of  4 chloride ions (ie. one radius from each corner chloride ion and one  diameter  which equals to two radii of  chloride ion ,  at the center of the face )

Thus  , when the length of  diameter is represented by d  and  radius of each chloride ion by  r ,

d   =  4r

applying Pythagorean theorem we get , 

a²    +   a²     =   d²  

Substituting the given value for  a  =   5.14 A^o   ( or =   0.514 nm ) in the above  equation ,

( 0.514 nm )²     +   ( 0.514 nm )²     =    d²   =   (4r) ²  =  16 r²   

^{{tex}\therefore{/tex}}  solving for  r   we get  ,

r    =   Sqrt [ {  (  0.514 nm )²    +   ( 0.514 nm )²    }  ]   / {tex}\sqrt16{/tex} 

     =  0.182 nm 

or , =  1.82 A^o   for  chloride ion 

according to henry's law the vapour pressure of gas is directly proportional to the concentration of gas or we can say the mole fraction of gas in the slution

At higher altitude the vapour pressure of oxygen decreases , which reults in the lowering of concentration of oxygen in our blood and results in anoxia

By arhenius equation

Considering  an equilibrium  established between the number of holes and the number of free electrons in a semiconductor at a particullar temperature.at  equilibrium let

, n_o    be the equilibrium free electron concentration in conduction  band  ,,  [ cm^-3 ]

p_o    be the equilibrium  hole concentration  in valency band  ,   [ cm^-3 ]

Certainly , then in intrinsic semiconductor :

.............n_o   =  p_o   =    n_i    

where,   n_i   represents  intrinsic - carrier concentration.

As  T  increases , then  n_i   increases .

As   E_g   increases   ,     n_i   decreases.

    We can consider this equilibria .........bond <---------->   e^-      +      h⁺        using an anology to chemical reaction  H₂O  <------------>   H⁺     +     OH^-    

Applying law of mass action ,

..................................................................K   =   [ e^- ] [  h⁺ ]  /  [ bond ]...................................................................&  K   =   [ H⁺ ] [ OH^- ]   /  [ H₂ O ]

........................................................................{tex}\approx{/tex}  exp [ - E_g  /  k T ]...................................................................................{tex}\approx{/tex}  [ - E  / k T ]

where , [ bonds ] is the concentration of unbroken bonds  ,  k is rate constant  &  E  , E_g   represents  respective  activation energies..

again , since [ bonds ]  >> n_o , p_o    

..................{tex}\therefore{/tex} [ bonds ]  =  constant

In general  , relatively few bonds are broken to form an electron hole , therefore, the number of bonds  are approximately constant .Hence ,

......................Number of holes   are equal to  number of free electrons

When intrinsic semiconductors like Si and Ge are doped with group-13 elements like B,Al or Ga which contains only three valence electrons, then the place where the fourth valence electron is missing is called electron hole.An electron from a neighbouring atom can come and fill the electron hole,but in doing so it would leave an electron hole at its majority position.

Since, the number of holes are more than the number of electrons,therefore they are called as p-tyoe semiconductors and holes are the majority carriers.

Solution : LiBr is Ionic, Brass is metallic, 

(NH₄)₃PO₄  is ionic.

Ans. Yes it is molecular 

Ans. A basic substance is capable of donating a lone pair towards forming a coordinate covalent bond. A secondary amine can donate its lone pair more easily, as the extra inductive effect stabilises the charge on the nitrogen atom.

Ans. S₈+ 48HNO₃ {tex}\to {/tex}8H₂SO₄ + 48NO₂ + 16H₂O

Ans.

• Molecular compounds are pure substances formed when atoms are linked together by sharing of electrons while ionic compounds are formed due to the transfer of electrons.
• Molecular compounds are made due to covalent bonding while ionic compounds are made due to ionic bonding.
• Molecular compounds are formed between two non-metals while ionic compounds are formed between metals and non-metals.
• Molecular compounds are poor electrical conductors while ionic compounds are good conductors.
• Molecular compounds can be in any physical state ‘“ solid, liquid, or gas. Ionic compounds are always solid and crystalline in appearance.

Ans. Because it is a stable cross linked polymer which attain a particular shape by heat and pressure. Chemical bonds formed are very strong. This is the reason why thermosetting plastics melt only the first time and become firm after that.

Ans. This happens because the energy difference between 5f, 6d and 7s subshell of the actinides is very small and hence electrons can be accommodated in any of them.