We know that to relate solutions of with the factors of molarity and volume, we can use the equation:

**
NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M

is the left side of the equation. Then we have:

We can then solve for

:

and

or

We now know that the total amount of volume of the 4.35 M solution will be
210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.
Answer:
The pH of the solution is 5.31.
Explanation:
Let "
is the dissociation of weak acid - HCN.
The dissociation reaction of HCN is as follows.

Initial C 0 0
Equilibrium c(1-
) c
c
Dissociation constant = 

In this case weak acids
is very small so, (1-
) is taken as 1.


From the given the concentration = 0.050 M
Substitute the given value.

![[H_{3}O^{+}]=c\alpha](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3Dc%5Calpha)
![[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D0.05%5Ctimes%209.8%5Ctimes%2010%5E%7B-4%7D%3D%204.9%5Ctimes10%5E%7B-6%7D)
![pH= -log[H_{3}O^{+}]](https://tex.z-dn.net/?f=pH%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
![=-log[4.9\times10^{-6}]](https://tex.z-dn.net/?f=%3D-log%5B4.9%5Ctimes10%5E%7B-6%7D%5D)

Therefore, The pH of the solution is 5.31.
The equation that we will use to solve these two questions is:
c = lambda * f where:
c is the speed f the wave
lambda is the wavelength of the wave
f is the frequency of the wave
Question 1:
we want to find f
c is assumed to be the speed of light = 3*10^8 m/sec
lambda = 5.6 * 10^10 cm = 560 * 10^6 meters
Substitute in the equation to get f as follows:
3*10^8 = 560*10^6*f
f = (3*10^8) / (560*10^6) = 0.5357 Hz
Question 2:
we want to find lambda
c is assumed to be the speed of light = 3*10^8 m/sec
f is given as 0.1096 Hz
Substitute in the equation to get lambda as follows:
3*10^8 = lambda * 0.1096
lambda = (3*10^8) / (0.1096) = 2.737226*10^9 meters
Although I’m not entirely sure what answer you’re looking for, I’d say because most scientific investigations require multiple trials of an experiment
3 is the only choice of these that is correct for this equation