Answer:
D
Explanation:
Cuz i got it wrong and this was the right answer : )
Answer:
The correct statements are A amine groups are fully protonated and can be described with the chemical formula NH3+ B carboxylic acid functional groups are de protonated and can be be described with the chemical formula COO- C Amine functional groups are positively.
Explanation:
If we study the biochemical structure of an amino acid we wil see that an amino group or -NH2 is present at one end and a carboxylic group or COOH is present at another end.
Now the fact the that when an amino acid exist as zwitterion it contain same number of positive charge as well as same number of negative charge.So during zwitterion formation the carboxylic acid or -COOH liberates a proton and exist as COO- whereas the amine group accepts that proton and exist as NH3+.
Beside this the amine group -NH2 after the formation of zwitterion gains a positive charge and exist as -NH3+.
What are the options? I can try to give you an answer if options are provided.
Atomic size gradually decreases from left to right across a period of elements.
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
- = 90.25 kJ/mol
- = 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
- =?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>