Class 12th
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New answer posted
a year agoContributor-Level 10
8.7 The following reactions are involved when Cr2+ and Fe2+ act as reducing
Cr2+ = Cr3+ ( E? Cr3+/ Cr2+ = - 0.41 V)
Fe2+⇒ Fe3+ (E? Fe3+/ Fe2+ = +0.77 V)
Since, Cr has less potential value, Cr2+ gets oxidised easily than Fe2+. Therefore, Cr2+ is a better reducing agent that Fe3+.
New answer posted
a year ago
Contributor-Level 10
(ii) Tertiary, because the Nitrogen atom is attached to the 3 carbon atoms.
(iii) Primary, because the nitrogen atom is attached to the only 1 carbon atom.
(iv) Secondary, because the nitrogen atom is attached to the only 1 carbon.
New answer posted
a year agoContributor-Level 10
5.26
Soap is sodium or potassium salts of fatty acids (long chain). Represented as RCOO–Na+. When dissolved in Water it dissociates in the RCOO– and Na+. In first ion there are two parts R and COO–. R is hydrophobic tail and COO– Hydrophilic head. RCOO– ions present at the surface with COO– group inside and R group outside.
Increasing concentration to critical micelle concentration anions are pulled inside the bulk and Form a spherical shape of hydrocarbon chain pointing towards the centre and COO– pointing outside this are called as micelle.
Cleansing action of soap: Soap molecules form micelle around oil droplets. So
New answer posted
a year agoContributor-Level 10
8.6 The oxidation state increases when an atom loses its Example: When Fe loses 2 electrons, its oxidation state becomes +2 from 0.
Oxygen (O) and fluorine (F) are very strong oxidizing agents. Both oxide and fluoride ions are highly electronegative and have a very small size, so they attract the electrons from metal atoms. Hence, they oxidize the metal to its the highest oxidation state.
New answer posted
a year agoContributor-Level 10
5.25
Emulsifiers form an interfacial film between the suspended particles and medium. It provide coating to every drop of suspended particle and prevent it from coagulating. And it remain suspended in medium. Ex- Proteins, gums, alcohols, Lampblack etc.
New answer posted
a year agoContributor-Level 10
8.5 The ionization energy increases due to the gradual filling of electrons in the d- subshells. The irregular variation of ionization energy is due to the fact that half-filled and completely filled subshells are more stable and have very high ionization energy.
In case of first ionization energy, Cr ( [Ar]3d54s1) attains the stable configuration (3d5) by losing one electron from s-subshell and hence, it has low ionization energy.
Whereas, Zn has high ionization energy because it has completely filled subshells and are very stable.
Second ionization energies are higher than the first since it becomes difficult to remove an electron
New answer posted
a year agoContributor-Level 10
5.25
Emulsions –The system which has Liquid as both dispersed phase and dispersion medium. There are two types of emulsions
Oil in Water-Ex-Milk
Water in oil-Ex-Butter.
New answer posted
a year agoContributor-Level 10
8.4 The E0 (M2+/M) value of a metal depends on the energy changes involved in the following reactions:
1. Sublimation energy: The energy needed to convert one mole of atoms from a solid state to gaseous
2. Ionization energy: The energy supplied to remove electrons from one mole of atoms, which are in the gaseous
3. Hydration energy: The energy emitted to hydrate one mole of
Now, copper has a high ionisation energy and low hydration energy. Hence, the E0 (M2+/M) value for copper is positive.
New answer posted
a year agoContributor-Level 10
5.23
(i) In this case Tyndall effect is observed i.e. scattering of light by the colloidal particles takes place and the path of light become visible.
(ii) The coagulation takes place in this case positive charge particles of Fe (OH)2 gets attached with Cl-. And of FeCl3 is obtained.
(iii) In this case coagulation will take place. After passing electric current through colloidal solution particle will get attracted towards oppositely charged and lose their charge and coagulate.
New answer posted
a year agoContributor-Level 10
8.3 Manganese (Z = 25) exhibits the largest number of oxidation states. This is because its electronic configuration is 3d54s2.
Because it has the maximum number of electrons (5 d electrons and 2s electrons) to easily lose and share.
Therefore, it can exhibit an oxidation state of +2 to +7. The compounds are as follows:
Mn (0) as Mn (s), Mn (II) as MnO, Mn (II, III), Mn etc.
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