Answer:
pH = 4.09
Explanation:
molarity of oxalic acid in the solution
= 0.1 x 25 / (25 + 35)
= 0.0417 M
molarity of NaOH in the solution
= 0.1 x 35 / (25 +35)
= 0.0583 M
H2C2O4 + NaOH -------------------> NaHC2O4 + H2O
0.0417 0.0583 0 0
0 0.0166 0.0417
now second acid -base titration
NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O
0.0417 0.0166 0 0
0.0251 0 0.0166 ---
now
pH = pKa2 + log [Na2C2O4 / NaHC2O4]
pH = 4.27 + log (0.0166 / 0.0251)
pH = 4.09
Answer:
Explanation:
From the given information:
A → B k₁
B → A k₂
B + C → D k₃
The rate law = ![\dfrac{d[D]}{dt}=k_3[B][C] --- (1)](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%3Dk_3%5BB%5D%5BC%5D%20---%20%281%29)
![\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C]](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BA%5D%20-k_2%5BB%5D%20-k_3%5BB%5D%5BC%5D)
Using steady-state approximation;
![\dfrac{d[B]}{dt}=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D0)
![k_1[A]-k_2[B]-k_3[B][C] = 0](https://tex.z-dn.net/?f=k_1%5BA%5D-k_2%5BB%5D-k_3%5BB%5D%5BC%5D%20%3D%200)
![[B] = \dfrac{k_1[A]}{k_2+k_3[C]}](https://tex.z-dn.net/?f=%5BB%5D%20%3D%20%5Cdfrac%7Bk_1%5BA%5D%7D%7Bk_2%2Bk_3%5BC%5D%7D)
From equation (1), we have:
![\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%3D%20%5Cdfrac%7Bk_3k_1%5BA%5D%5BC%5D%7D%7Bk_2%2Bk_3%5BC%5D%7D%7D)
when the pressure is high;
k₂ << k₃[C]
![\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \ \text{first order}](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%20%5Cdfrac%7Bk_3k_1%5BA%5D%5BC%5D%7D%7Bk_3%5BC%5D%7D%3D%20k_1A%20%5C%20%5C%20%20%5Ctext%7Bfirst%20order%7D)
k₂ >> k₃[C]
![\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \ \text{second order}](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%20%5Cdfrac%7Bk_3k_1%5BA%5D%5BC%5D%7D%7Bk_2%7D%3D%20%5Cdfrac%7Bk_1k_3%7D%7Bk_2%7D%5BA%5D%5BC%5D%20%5C%20%5C%20%20%5Ctext%7Bsecond%20order%7D)
Answer:
Mass = 29.23 g
Explanation:
Given data:
Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)
Molarity of solution = 0.227 M
Mass of solute in gram = ?
Solution:
Molarity = number of moles / volume in L
By putting values,
0.227 M = number of moles / 0.8142 L
Number of moles = 0.227 M × 0.8142 L
Number of moles = 0.184 mol
Mass in gram:
Mass = number of moles × molar mass
Molar mass of calcium acetate = 158.17 g/mol
Mass = 0.184 mol × 158.17 g/mol
Mass = 29.23 g
Answer:
<em>1 mole is equal to 1 moles NaOH, or 39.99711 grams.</em>
Explanation:
<em>Hope this helps have a nice day :)</em>
A Product is a new substance formed in a chemical reaction
