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Answer:
k = 1.3 x 10⁻³ s⁻¹
Explanation:
For a first order reaction the integrated rate law is
Ln [A]t/[A]₀ = - kt
where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.
We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus
Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt
- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹
Which element requires the least amount of
energy to remove the most loosely held electron
from a gaseous atom in the ground state?
<h3>Answer-</h3><h3>Na</h3>
<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
The correct answer would be the last option. A double displacement type of reaction involves the switching of places the cations and anions accordingly. The given reaction is erroneous since in the product side the anions and cations are being paired which would not make sense. The correct reaction should be
4NaBr + Co(SO3)2 yields <span>CoBr4 + 2Na2SO3</span>
