Answer:
vp jokhimon vf dpp gl fl vk hggjuvg7vvohohohohojj
Conjugate base of Propanoic acid (
is propanoate where -COOH group gets converted to -CO
. The structure of conjugate base of Propanoic acid is shown in the diagram.
The
above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.
=
+ log
=4.9+log
=5.85
As 90% conjugate base is present, so propanoic acid present 10%.
"The reaction will absorb energy" is the best conclusion according to the energy diagram of the chemical reaction.
<u>Option: B</u>
<u>Explanation:</u>
The chemical bonds in the reactions are broken and formed as per process and contributed by three major steps: reactants, transition phase and product formation. Here transition phase is in equilibrium stage drived by activation energy, where bond is partially formed and partially broken, located at higher energy level then the starters.
The reactant's energy level is less relative to the products as seen in the endothermic reactions' energy diagram, which depicts that the products are less balanced than reactants. Here when the reaction is forced to the forward direction, then it direct towards the more unbalance entities. As energy is absorbed in the endothermic reaction from surrounding, thus the enthalpy change (ΔH) for the reaction is positive.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:

Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 



Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L

Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.