Answer:
electric is your answer :)
Explanation:
Edge STEM instruction/assignment perhaps???
The oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.
<h3>What is an oxidizing and reducing agent?</h3>
An oxidizing agent is any substance that oxidizes, or receives electrons from another substance and as a result, becoming reduced.
On the other hand, a reducing agent is any substance that reduces or donates electrons to another and as a result becomes oxidized.
According to this reaction; H2S(aq) + Cl2(g) -> S(s) + 2HCI (aq)
- H2S accepts electrons from Cl2 and becomes reduced to S
- Cl2 donates electrons to H2S and becomes oxidized to HCl
Therefore, the oxidizing and reducing agent in the above redox reaction are hydrogen sulphide (H2S) and Chlorine (Cl) respectively.
Learn more about oxidizing agent at: brainly.com/question/10547418
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Answer:
Farthest from the carbonyl carbon.
Explanation:
Reference carbon that determined the absolute D and L configuration is located farthest from the carbonyl carbon.
In other words, reference carbon is that assymentric carbon which is located farthest from the carbolyl carbon and has configuration similar to D- or L-glyceraldehyde isomers.
D and L configuration is decided by the direction of -OH group attached to the reference carbon.
In L-isomer, -OH group is attached to the left side of the reference carbon and in D-isomer, -OH group is attached to the right side of the reference carbon.
Best Answer: <span>(a)
8.9 x 10^-7 = x^2 / 0.15-x
x = [OH-] = 0.00037 M
pOH = 3.4
pH = 14 - 3.4 = 10.6
(b)
Ka = Kw/Kb = 5.6 x 10^-10 = x^2 / 0.20-x
x = [H+] = 0.000011 M
pH = 5.0</span>
1.9 mol H2.
3.8 mol HCl * (1 mol H2 / 2 mol HCl) = 1.9 mol H2
