Answer:
The procedure you will use in this exercise exploits the difference in acidity and solubility just described.
(a) you will dissolve your unknown in ethyl acetate (an organic solvent). All of the possible compounds are soluble in ethyl acetate.
(b) you will extract with sodium bicarbonate to remove any carboxylic acid that is present.
(c) you will extract with sodium hydroxide to remove any phenol that is present.
(d) you will acidify both of the resulting aqueous solutions to cause any compounds that were extracted to precipitate.
Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!
Answer:
HNO₂
Explanation:
An acid is a proton donor; a base is a proton acceptor.
Thus, NO₂⁻ is the base, because it accepts a proton from the water.
H₂O is the acid, because it donates a proton to the nitrite ion.
The conjugate base is what's left after the acid has given up its proton.
The conjugate acid is what's formed when the base has accepted a proton.
NO₂⁻/HNO₂ make one conjugate acid/base pair, and H₂O/OH⁻ are the other conjugate acid/base pair.
NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻
base acid conj. conj.
acid base
I dont know your question but that is true
Answer:
Freezing point is -2.81°C
Explanation:
34g/342gmol^-1 = 0.0994mol
n = m/mr
Molarity= 0.994/ 0.66 = 1.51M
◇T = -i × m ×Kf
Where ◇T is freezing depression
i= Vant Hoff factor
m = molarity
Kf = freezing content = 1.
860kgmol^-1
◇T =-1 × 1.51 × 1.860 = - 2.81°C
