Here ya go if you need the link we’re I found this answer key https://studylib.net/doc/8708211/exam-2---chemistry
Answer:
pH = 12.52
Explanation:
Given that,
The [H+] concentration is
.
We need to find its pH.
We know that, the definition of pH is as follows :
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Put all the values,
![pH=-log[3\times 10^{-13}]\\\\pH=12.52](https://tex.z-dn.net/?f=pH%3D-log%5B3%5Ctimes%2010%5E%7B-13%7D%5D%5C%5C%5C%5CpH%3D12.52)
So, the pH is 12.52.
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
ANSWER IS CONDUCTION. HOPE THIS HELPED!
Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

Finally, the resulting molarity in 30.8 mL (0.0308 L):

Regards.