I'm very confused Valine is an amino acid that can take on many protonated and unprotonated forms. It has a weakly acidcarboxylic acid group (pK, = 2.32) and a weakly basic amine group (pKa = 9.62).Identify which of the forms below would likely exist at the given pH values.H2N*H3NОНStructure AStructure BH2N*H3N.O.ОНStructure CStructure DWhen the pH = 1.00, [ Select ]predominates.When the pH = 5.00, [ Select ]predominates.When the pH = 12.00, [ Select]* predominates. A 6.8 L sample of helium gas stored at 25 °C and 1.00 atm pressure is transferred to a 2.4 L tank andmaintained at a pressure of 2.37 atm. What temperature (in °C) is needed to maintain this pressure?

Interpretation: The pKa value of carboxylic acid group in value is low. This means that it is highly acidic. It readily donated its proton when a base is available. So, at high pH values the carboxylic acid gets deprotonated. The pKa value of amine group in value is high. This means that it is weakly basic. It will not accept the proton readily when an acid is available. At pH values near to neutral it will not get protonated but at very low pH values it will get protonated. When pH = 1.00: pH value is very low. This means that it is highly acidic in nature. So, the amine group gets protonated and the carboxylic acid group remains unaltered. Structure- D. When pH = 5.00: pH value is near to neutral and slightly acidic. As the amine group is a weak base it will not get protonated easily as the acidic condition is so weak. The carboxylic group will be unaltered as there is no basic condition to accept the proton. Structure- A. When pH = 12.00: pH value is very high. This means that it is highly basic in nature. So, the carboxylic acid group will get deprotonated. The amine group remains unaltered. Structure- C.Initial volume of Helium gas (V1) = 6.8 L Initial temperature of Helium gas (T1) = 25°C = 25+273.15=298.15K Initial pressure of Helium gas (P1) = 1.00atm Final volume of Helium gas (V2) = 2.4 L Final pressure of Helium gas (P2) = 2.37atm Final temperature of Helium gas (T2) = x According to the Ideal gas equation: P×V=n×R×TP×VT=n×R As the number of moles of the remains constant, the above equation can be written as: P1×V1T1=P2×V2T2 T2=P2×V2×T1P1×V1Substituting the values in the above equation T2=2.37 atm×2.4 L×298.15K1.00 atm×6.8 L T2=249.39K Temperature in (°C) =249.39 K - 273.15=-23.76°C Therefore, the temperature needed to maintain this pressure is -23.76°C