From the calculations, the pH of the final solution is 9.04.
<h3>What is the pH of the buffer?</h3>
We can use the Henderson Hasselbach equation to obtain the final pH of the solution in terms of the pKb and the base concentration.
Number of moles of salt = 250/1000 L * 0.5 M = 0.125 moles
Number of moles of base = 150/1000 L * 0.5 M = 0.075 moles
Total volume of solution = 250ml + 150ml = 400ml or 0.4 L
Molarity of base = 0.075 moles/ 0.4 L = 0.1875 M
Molarity of salt = 0.125 moles/ 0.4 L = 0.3125 M
pOH = pKb + log[salt/base]
pKb = -log(1.8 x 10^-5) = 4.74
pOH = 4.74 + log[0.3125/0.1875 ]
pOH = 4.96
pH = 14- 4.96
pH = 9.04
Learn more about pH:brainly.com/question/15289741
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the element would be Iodine (I) bc it has 53 protons
<span>Heat capacity of an object, is the amount of heat energy or thermal energy (unit: Joule) needed to raise the temperature of the object by 1 degree celsius. Unit of heat capacity is J/°C
Larger object will surely need larger amount of thermal energy to raise its temperature. If you compare 1 litre of water with 0.5 litre of water, the 1L water will have two times the heat capacity.
It will be more useful to compare specific heat capacity, because then it is the amount of heat energy or thermal energy (unit: Joule) needed to raise the temperature of 1 unit mass of the object by 1 degree celsius. You can then compare between 1 unit mass of water and 1 unit mass of iron.
Water has higher specific heat capacity than iron, meaning that you need more energy to heat up 1kg of water, then to heat up 1kg of iron.
The unit will then be J/(kg °C) or J/(g °C).
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Answer:
D For Doras Demolishing Dub Dune Is Dungenes
Answer: C, B, A
Explanation:
C is the initial solution, because naoh has not been added yet
B is the midpoint of the titration. naoh has been added to the solution, but it has not fully reacted yet. You can tell that this one is the midpoint because there is still HF- in the diagram, which is not one of the products formed in the reaction.
A is the endpoint because the diagram shows that all products are fully formed. There is an increased amount of Na+ and H2O in the diagram and no HF- left.
PS. I tested this answer on the concentration of acetic acid post-lab (from mcgraw hill) earlier today and it said this was the right answer :) hope this helps
