D represents ion-dipole forces that are stronger than the force C.
Explanation:
D represents the ion-dipole force.
C represents the H-bonding forces.
ion-dipole force is a force that is due to electrostatic attraction and has a dipole between an ion and a neutral molecule.
It is electrostatic in nature.
A hydrogen bond is the force between the hydrogen with the electro negative atom of one molecule, to electro negative atom of some other molecule. such as: O, F, N
Ion dipole force is stronger than the H-bonding.
31 protons 31 electrons and 39 nutrons
Answer: 4.25
Explanation: CsOH=Cs+
pOH= -log10
= -log10(5.6*10^-5)
= -log10(5.6*10^-5)=-(-4.25)
=4.25
In titration, the moles of acid equal moles of base. You were given that 22.75ml of 0.215M NaOH is used, so calculate the number of moles of that base the experiment used in total. After that because you know mol base = mol acid, whatever amount of base you use must be the total amount of acid present in the solution. You were given the volume of the acid, and you have just found the total mols of acid. Using these two information, solve for the concentration. And one more thing, even though I'm pretty sure it won't affect your answer, you should always convert things to the proper units. Since the concentration we're talking about in this problem is molarity, which has the unit mol/L, you should always have all of your numbers in these units. It just make it simpler and will not confuse you
<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:
Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:
<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'
The expression of 
for above equation follows:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![[HI]_{eq}=2x=(2\times 0.079)=0.158M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.079%29%3D0.158M)
![[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D%280.1-x%29%3D%280.1-0.079%29%3D0.0210M)
![[I_2]_{eq}=0.0210M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.0210M)
Putting values in above expression, we get:
Hence, the value of equilibrium constant for the given reaction is 56.61
