The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.
<h3>What is equilibrium?</h3>
Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
A very high value of K indicates that at equilibrium most of the reactants are converted into products.
The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.
When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.
This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.
Hence, option A is correct.
Learn more about the equilibrium here:
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Answer:
pI = 6.16
Explanation:
The pI is given by the average of the pKas that are involved. In this case,
Pka of carboxylic acid was given as 2.72 and that of the Amino group was given as 9.60. the average would then be ½(2.72+9.60)
= 6.16
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Hope this help
Answer:
237.8L of water would need to be added.
Explanation:
The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).
Then figure out what information you have and what is being found. In this case:
M1 = 54.7 M
V1 = 1092 mL = 1.092 L
M2 = 0.25 M
V2 = unknown
Then solve the equation for whatever you are trying to find.
M1V1=M2V2
V2=M1V1/M2
Now you need to plug everything in.
V2=(54.7M*1.091L)/0.25M
V2=238.93L
That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.
238.93L - 1.091L = 237.8L
Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.
I hope this helps. Let me know if anything is unclear.
