D.
The volume of water in a graduated cylinder
should be read at the bottom of the curve of the meniscus.
Riley can either change the surface area of the object or can change the slipperiness of the material.
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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)
![rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%20%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
(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} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}](https://tex.z-dn.net/?f=%20%2B%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20-%5Cfrac%7B2%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5B-1.61%20%5Ccdot%2010%5E%7B-5%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!
Answer:
6,04x10⁻³M
Explanation:
For the reaction:
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
The precipitate of Cu(s) weights 96,0 mg. In moles:
Moles of Cu(s):
0,096g×(1mol/63,546g) = 1,51x10⁻³ moles of Cu(s). If you see the balanced equation 1 mole of CuSO₄ produce 1 mole of Cu(s). That means moles of CuSO₄ are the same of Cu(s), <em>1,51x10⁻³ moles of CuSO₄</em>
As volume of the solution is 250 mL, 0,250L, the molar concentration of the original solution is:
1,51x10⁻³ moles of CuSO₄ / 0,250L = <em>6,04x10⁻³M</em>
I hope it helps!
