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olasank [31]
2 years ago
9

Ants live in the hollow thorns of an acacia tree. The tree provides food and shelter for the ants. The ants protect the tree fro

m herbivores that might otherwise eat the tree.
Which type of relationship does this represent?
A. mutualism
B. parasitism
C. commensalism
D. predator/prey
Chemistry
2 answers:
AfilCa [17]2 years ago
7 0
A. Mutualism


Mutualism: Both organism benefit

Commensalism: One organism benefits. One organism is unaffected

Parasitism: One organism benefits. One organism is harmed
tatuchka [14]2 years ago
5 0
Answer:
A. mutualism

explanation:
mutualism is the mutual dependence that is necessary for well being. the runner up answer would be commensalism. commensalism is the association between two organisms in which one benefits and the other derives neither benefit nor harm. which is not true because both benefit.
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g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
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Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
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