Answer:
Cl2 +2Kl ----->2KCL +I2
Explanation:
_Cl2 + __KI → __KCI + __I2
Cl2 +2Kl ----->2KCL +I2
Answer:
The answer to your questions is Cm = 25.5 J/mol°C
Explanation:
Data
Heat capacity = 0.390 J/g°C
Molar heat capacity = ?
Process
1.- Look for the atomic number of Zinc
Z = 65.4 g/mol
2.- Convert heat capacity to molar heat capacity
(0.390 J/g°C)(65.4 g/mol)
- Simplify and result
Cm = 25.5 J/mol°C
Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :

where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:



Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
D, It is a flow of protons, is the best answer. Electricity is the flow of electrons, not protons.
Answer:
We can do the nitration of benzene by treating the benzene with a mixture of nitric acid and sulphuric acid by not extending the temperature of 50°C
Explanation:
Nitration of benzene takes place by treating the benzene with a mixture of nitric acid and sulphuric acid at low temperatures such as the temperatures below 50°C
The nitration of benzene takes place through electrophilic substitution reaction
In this reaction the electrophile is nitronium ion (NO2+) which performs an electrophilic substitution reaction on the benzene ring and during the reaction an intermediate will also be formed in which there will be positive charge distributed in the benzene
These electrophile is generated when nitric acid is treated with sulphuric acid
As nitric acid is a strong oxidising agent, here in this case the oxidation state of nitrogen will change from +5 to +3
The reactions regarding the nitration of benzene is present in the file attached