K, P, K, K, P, K, K, P, K, P. If it is moving, it is kinetic, if it isn't, it's potential. the sugar one is a little tricky using that method though, because we generally consider this in terms of spacial movement, but sugar holds energy which is later released by your body to allow you to move.the chemical bonds have potential energy because they release energy when broken.
<span>The correct answer is that an ionic bond forms between charged particles. To form this bond, the particles transfer valence electrons (those in the outermost orbit). Specifically, in ionic bonding, the metal atom loses its electrons (thus becoming positive) and the nonmetal atom gains electrons (thus becoming negative).</span>
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:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
