Answer:electrons will be equally attracted to both
Explanation:
e2020
Answer:
0.529
Explanation:
Let's consider the reaction A → Products
Since the units of the rate constant are s⁻1, this is a first-order reaction with respect to A.
We can find the concentration of A at a certain time t (![[A]_{t}](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D)
) using the following expression.
![[A]_{t}=[A]_{0}.e^{-k\times t}](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D%3D%5BA%5D_%7B0%7D.e%5E%7B-k%5Ctimes%20t%7D)
where,
[A]₀: initial concentration of A
k: rate constant
![[A]_{t}=0.548M.e^{-3.6\times 10^{-4}s^{-1}\times 99.2s }](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D%3D0.548M.e%5E%7B-3.6%5Ctimes%2010%5E%7B-4%7Ds%5E%7B-1%7D%5Ctimes%2099.2s%20%7D)
![[A]_{t}=0.529 M](https://tex.z-dn.net/?f=%5BA%5D_%7Bt%7D%3D0.529%20M)
<em>V= 110mL = 110cm³ = 0,11dm³</em>
<em>C = 1,244 mol/L = 1,244 mol/dm³</em>
C = n/V
n = 1,244×0,11
<u>n = 0,13684 moles</u>
<em>mCa(OH)₂ = 74 g/mol</em>
1 mole Ca(OH)₂ ------------ 74g
0,13684 ---------------------- X
X = 74×0,13684
<u>X = 10,12616g</u>
:)
Answer:
Coefficient = 1.58
Exponent = - 5
Explanation:
pH = 2.95
Molar concentration = 0.0796M
Ka = [H+]^2 / [HA]
Ka = [H+]^2 / 0.0796
Therefore ;
[H+] = 10^-2.95
[H+] = 0.0011220 = 1.122 × 10^-3
Ka = [H+] / molar concentration
Ka = [1.122 × 10^-3]^2 / 0.0796
Ka = (1.258884 × 10^-6) / 0.0796
Ka = 15.815 × 10^-6
Ka = 1.58 × 10^-5
Coefficient = 1.58
Exponent = - 5
