Answer:
The ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .
Explanation:
HC2H3O2 is CH₃⁻ COOH, which is also known as Acetic acid.
IUPAC name of this compound is Ethanoic acid.
Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.
Given that, equivalence point was reached when 20.0mL of NaOH is added.
let the normality of acetic acid is N₁ and that of NaOH is N₂.
volume of acetic acid is V₁ and that of NaOH is V₂.
Equivalence point occurs when, N₁ × V₁ = N₂ × V₂.
⇒ N₁ × V₁ = N₂ × 20.
after the addition of 5.0mL of NaOH(aq), remaining N₁ × V° = N₂ × (20 - 5).
= N₂ × 15.
after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).
= N₂ × 19.
⇒ V° : Vˣ = 15 : 19 .
⇒
M(Mn(ClO3)3)=(54.938)+(35.45x3)+(15.999x9)
M(Mn(ClO3)3)=305.279 g/mol
Answer:
(a) -0.00017 M/s;
(b) 0.00034 M/s
Explanation:
(a) Rate of a reaction is defined as change in molarity in a unit time, that is:
![r = \frac{\Delta c}{\Delta t}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5CDelta%20c%7D%7B%5CDelta%20t%7D)
Given the following reaction:
![2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)](https://tex.z-dn.net/?f=2%20N_2O_5%20%28g%29%5Crightleftharpoons%204%20NO_2%20%28g%29%20%2B%20O_2%20%28g%29)
We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:
![r = -\frac{\Delta [N_2O_5]}{2 \Delta t}](https://tex.z-dn.net/?f=r%20%3D%20-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D)
Reaction rate is also equal to the rate of formation of products divided by their coefficients:
![r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BO_2%5D%7D%7B%5CDelta%20t%7D)
Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:
![r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s](https://tex.z-dn.net/?f=r_%7BN_2O_5%7D%20%3D%20%5Cfrac%7B0.066%20M%20-%200.100%20M%7D%7B200.00%20s%20-%200.00%20s%7D%20%3D%20-0.00017%20M%2Fs)
(b) Using the relationship derived previously, we know that:
![-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D)
Rate of appearance of nitrogen dioxide is given by:
![r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}](https://tex.z-dn.net/?f=r_%7BNO_2%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D)
Which is obtained from the equation:
![-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B4%20%5CDelta%20t%7D)
If we multiply both sides by 4, that is:
![-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B4%20%5CDelta%20%5BN_2O_5%5D%7D%7B2%20%5CDelta%20t%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D)
This yields:
[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]
Here is my answer. I hope this is helpful.