Due to common ion effect, addition of 1.0mL of 4MNaCl will cause some of the Cu+ and Cl− ions combine to form CuCl(s) because the Ksp will be lower than 1.6×10−7.
<h3>What is solubility product?</h3>
The term solubility product refers to the equilibrium constant that is set up when an ionic substance dissolve in water. Since the solution already contains chloride ions, addition of more chloride ions from NaCl will cause more CuCl(s) to separate from solution.
Hence, when 1.0mL of 4MNaCl is added to the solution, some of the Cu+ and Cl− ions combine to form CuCl(s) because the Ksp will be lower than 1.6×10−7.
Learn more about solubility product: brainly.com/question/857770
Lead is most likely the best conductor
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. 
Major Species at B:
1.
2. 
Major Species at C:
1.
Major Species at D:
1.
2. 
Major Species at E:
1.
Major Species at F:
1.
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK = 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H]/[H2]
pH = pK = 10.33
I think the correct answer from the choices listed above is the first option. It would be several gold atoms that will form <span>a metallic bond. It is the most logical choice since it is a metal while the others are nonmetal and a combination of metal and nonmetal. Hope this answers the question. Have a nice day.</span>
Answer:
5511 J
Explanation:
Applying,
Q = Cm.................... Equation 1
Where Q = amount of heat required to convert ice, C = Heat of fusion of water, m = mass of ice
From the question,
Given: C = 334 J/g, m = 16.5 g
Substitute these values into equation 2
Q = 334(16.5)
Q = 5511 J
Hence, the amount of heat required is 5511 J
