Hope you understand how to work out those types of questions now xD ;)
The question is incomplete, the complete question is:
A chemist prepares a solution of vanadium (III) chloride (VCl3) by measuring out 0.40g of VCl3 into a 50.mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl− anions in the chemist's solution. Be sure your answer is rounded to the correct number of significant digits.
Answer:
0.153M of anions
Explanation:
First we calculate the concentration of the solution. From m/M= CV
m=given mass, M= molar mass, C =concentration of solution, V= volume of solution
Molar mass of compound= 51 + 3(35.5)= 157.5gmol-1
0.4g/157.5gmol-1= C×50/1000
C= 2.54×10-3/0.05= 0.051M
But 1 mole of VCl3 contains 3 moles of anions
Therefore, 0.051M will contain 3×0.051M of anions= 0.153M of anions
Answer:
Hydration
Explanation:
In water, ions are surrounded by a sphere of water molecules called a hydration shell. The process of forming this shell is called hydration.
I cannot come up with a reaction which you can convert directly from Al to H2O.
But you can convert from Al2O3 to H2O by adding HCL solvent.
Answer:
(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt
(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls
Explanation:
By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>
aX → bY (1)
![rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%3D%20-%5Cfrac%7B1%7D%7Ba%7D%20%5Cfrac%7B%5CDelta%5BX%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20%2B%5Cfrac%7B1%7D%7Bb%7D%20%5Cfrac%7B%5CDelta%5BY%5D%7D%7B%20%5CDelta%20t%7D%20)
<em>where, a and b are the coefficients of de reactant X and product Y, respectively. </em>
(a) Based on the definition above, we can express the rate of reaction (2) as follows:
3O₂(g) → 2O₃(g) (2)
(3)
(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:
![rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
![\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%201.07%20%5Ccdot%2010%5E%7B-5%7D%20%5Cfrac%7Bmol%7D%7BLs%7D%20)
So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.
Have a nice day!