Answer:
In doubling the concentration of the alkyl halide, the reaction rate also increases two-fold. However, doubling the concentration of the nucleophile does not in any way alter the reaction rate. Thus, the reaction rate is proportional only to the alkyl halide's concentration.
Answer: 94.13 L
Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.
To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.
A is the answer
Hope it helps :)
Explanations:
<u>Question</u> <u>1:</u> Lithium in 20.00+ g is C. or D., but 25.00+ g is D. which means this is the correct option.
I am unsure of <u>Question</u> <u>2</u>. I don't think it is mole though.
<u>Question</u> <u>3:</u> Boron in 25.00-30.00 g is B. or D., but 25.00 g would be C.
<u>Question</u> <u>4:</u> 2.393 x 1024 atoms of Oxygen is 63.58 mole O. I don't know for sure, but I think this is correct.
<u><em>I am NOT professional. There is a chance I am incorrect. Please reply to me if I've made a mistake.</em></u>
Answer:
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution is 3.72.
Explanation:
The 
The value of the dissociation constant = 
![pK_a=-\log[K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%5BK_a%5D)

Initial concentration of the acetic acid = [HAc] =c = 0.00225
Degree of dissociation = α

Initially
c
At equilibrium ;
(c-cα) cα cα
The expression of dissociation constant is given as:
![K_a=\frac{[H^+][Ac^-]}{[HAc]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BAc%5E-%5D%7D%7B%5BHAc%5D%7D)



Solving for α:
α = 0.08448
The degree of dissociation of acetic acid is 0.08448.
![[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%20%3D%200.00225M%5Ctimes%200.08448%3D0.0001901%20M)
The pH of the solution ;
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![=-\log[0.0001901 M]=3.72](https://tex.z-dn.net/?f=%3D-%5Clog%5B0.0001901%20M%5D%3D3.72)