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New answer posted
11 months agoContributor-Level 10
4.6 Given-
t1/2 = 60 mins
Using the formula for half life, t1/2 = 0.693/k, we get,
k= 0.693 / t1/2
k= 0.693 / 60
∴k = 1.155 * 10-2 min-1
The rate constant of the reaction, k is 1.155 * 10-2 min-1
New answer posted
11 months agoContributor-Level 10
(i) Let Oxidation of Co be x and charge on the complex is given as + 2
H2O has Oxidation Number: 0
CN has Oxidation Number: -1
en has Oxidation Number :0
(ii) Let Oxidation number of Co be x and charge on the complex is given as + 1
Br has Oxidation number: 1
en has oxidation number : 0
(iii) Let Oxidation number of Pt be x and charge on the complex is given as -2
Cl has oxidation number : -1
(iv) This complex can also be seen as [Fe (CN)6]3-
Let Oxidation number of Fe be x and charge given on the complex is given as -3
CN has oxidation number : -1
(v) Let Oxidation number of Cr be x and charge given on the complex is given
New answer posted
11 months agoContributor-Level 10
4.5 Given-
Rate constant, k = 1.15 * 10-3 s-1
innitial quantity R0 = 5g
Final quantity, R = 3 g
According to the formula of first order reaction
K = 2.303/ t log R0 / R
1.15 X 10-3 = 2.303 /t log 5/3
t = 2.303 / 1.15 X 10-3 log 5/3
t = 443.8 s
t = 4.438 * 102 s
The time taken for 5g of the reactant to reduce to 3 g is 4.438 * 102 s
New answer posted
11 months agoContributor-Level 10
4.4 Let the reaction be X→Y
As this reaction follows second order kinetics, the rate of the reaction will be,
Rate = k (X)2 → Equation 1
Since the concentration of X is increased three times, Equation 1 will become,
Rate = k (3X)2
= k * 9 (X)2
∴ The rate of formation of Y will become 9 times.
Thus, the rate of formation of Y will become 9 times
New answer posted
11 months agoContributor-Level 10
Ligands are the neutral or negatively charged entities surrounding the central metal atom of the coordination complex which possesses at least one unshared pair of electrons.
Based on the number of donor sites of these ligands, Ligands are classified as:
Unidentate ligands: These Ligands which have only one donor site are called unidentate ligands.
Example: F-, Cl – etc.
Didentate ligands: These Ligands which have only two donor site are called didentate ligands.
Example: Ethane-1,2-diamine, Oxalate ion etc.
Ambidentate ligands: These ligands which can attach them with the central metal atom by two different atoms are called as ambidentate
New answer posted
11 months agoContributor-Level 10
14.33
The different types of RNA found in the cell are listed below:-
(i) Messenger RNA (m-RNA)
It carries the genetic message code from the DNA to ribosomes. It is produced by the DNA; m-RNA is also single stranded and constitutes about 15% of total RNA.
(ii) Ribosomal RNA (r-RNA)
It is found in the ribosomes and it is usually associated with protein to form the ribosomes. It is synthesised in the nucleus by DNA. It is single stranded, comprising about 80% of total RNA. It is metabolically stable.
(iii) Transfer RNA (t-RNA)
It is synthesised in nucleus by DNA. It is also called soluble RNA. It is single stranded. There are 20
New answer posted
11 months agoContributor-Level 10
(1) Coordination Entity: Coordination entity is a charged entity having positive or negative charge in which the central atom is surrounded by molecules which may be neutral/negatively charged called Ligands
Examples:
i. Cationic Complexes: [Cu (H2O)6]2+, [Al (H2O)6]3+
ii. Anionic Complexes: [CuCl4]2-, [Al (H2O)2 (OH)4]-
iii. Neutral Complexes: [Co (NH3)4 Cl2], [Ni (CO)4]
(2) Ligands: Ligands are the neutral or negatively charged entities surrounding the central metal atom of the coordination complex which possesses at least one unshared pair of electrons
Example: F-, Cl-, Br-, I-, H20, and NH3
(3) Coordination Number: Coordina
New answer posted
11 months agoContributor-Level 10
14.32
The structural difference between DNA and RNA are as follows:-
The functional difference between DNA and RNA are as follows:-
New answer posted
11 months agoContributor-Level 10
The reaction is given below:
FeSO4 + (NH4)2SO4 + 6H2O → FeSO4 (NH4)2SO4.6H2O (Mohr Salt)
FeSO4, when reacted with (NH4)SO4, does not form any complex whereas they form a double salt, FeSO4. (NH4)2SO4.6H2O - (Mohr salt) which dissociates into ions in the solution. So, it gives the test of Fe2+ ions.
CuSO4 + 4NH3 + 5H2O→ [Cu (NH3)4SO4].5H2O
CuSO4 solution when mixed with aqueous ammonia in 1: 4 molar ratio forms a complex with formula [Cu (NH3)]SO4 in which the complex ion, [Cu (NH3)4]2+ does not dissociate to give Cu2+ ions. Therefore, it does not give the tests of the Cu2+ ion.
New answer posted
11 months agoContributor-Level 10
Bonding in coordination compounds in terms of Werner's postulates is explained as:
(a) Metals can show two types of valencies which are Primary valency and Secondary
1. Primary Valency: Primary Valency shows Oxidation Primary valencies are ionizable.
2. Secondary Valency: Secondary Valency shows coordination These are non-ionizable.
(b) Both Primary and secondary valency of the metal are to be satisfied which is done by negative ions in case of primary valency and negative or neutral species in case of secondary
(c) Metals have a fixed number of secondary valencies/ Coordination number around the central atom, these secondary
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