Response:
Unfiltered polluted water appears unclear and has a yellowish tint. The filtered polluted water is clear, albeit with a slight yellow hue. The pH levels exceed the tolerable range for organism survival. From these observations, it can be inferred that even if water seems clean, it may not be safe and could still pose risks to living organisms.
Greetings!
To tackle this question, we will apply the
Henderson-Hasselbach equation and solve for the molar ratio. It’s essential to obtain the pKa value for Acetic Acid, which is listed in reference tables as
4.76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
Thus, the mole ratio of CH₃COOH to CH₃COONa is
5.75Wishing you a wonderful day!
Answer:
Mass of Fe = 32.55 g
Explanation:
The molar mass of a single Fe atom is 55.845 g/mol
This means that
1 mole of Fe equals = 55.845 g ..............(1)
According to the mole concept,[ [TAG_24]]
1 mole of Fe contains = 
atom..........(2)
If we substitute 1 mole in equation (2) with 55.845 g, it becomes:
55.845 g of Fe contains = atoms
Reversing the equation yields
atoms of Fe correspond to 55.845 g
1 atom of Fe corresponds to 
g
one atom corresponds to
On computation,
one atom corresponds to 32.55 g of Fe
Mass of Fe = 32.55 g
<span>Response:
A 1.00 L solution that includes 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en).
Contains
0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine.
Using the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with double that amount of en = 0.000600 mol of en.
Thus, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted leaves 0.00180 mol en unreacted.
According to the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to yield an equivalent of 0.000300 moles of Cu(en)2^2+
The formation constant Kf for Cu(en)2^2+ is 1x10^20.
Therefore,
1 Cu+2 and 2 en --> Cu(en)2^2+
Kf = [Cu(en)2^2+] / [Cu+2] [en]^2
1x10^20 = [0.000300] / [Cu+2] [0.00180 ]^2
Solving for [Cu+2] gives [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Thus, Cu+2 = 9.26 e-19 Molar.
Since Kf only has 1 significant figure, round that to 9 X 10^-19 Molar Cu+2.</span>
For the first-order decomposition, the equation is: ln(x0 / x) = kt. At t = 200, x = 0.0300 M, we have ln(x0 / 0.03) = 200k. At t = 400, when x = 0.0200 M, we utilize ln(x0 / 0.02) = 400k. By multiplying the first equation by 2, we get 2ln(x0 / 0.03) = 400k, which aligns with the second equation, leading us to conclude that 2ln(x0 / 0.03) = ln(x0 / 0.02). This suggests (x0 / 0.03)^2 = x0 / 0.02, allowing us to find x0 = 0.045 M as the initial concentration. Plugging this back into the first equation yields: ln(0.045 / 0.03) = 200k, from which it follows that k = 0.0020273 (rate constant). The half-life can be calculated with x = 0.5x0: ln(x0 / 0.5x0) = 0.0020273t, resulting in ln(2) = 0.0020273t, which simplifies to t = 341.90 minutes (half-life).
