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Afina-wow [57]
3 years ago
7

Helppppp meeeeeeeeeeeeeeeeeeeeeeeeeeeee ill mark brainlist

Chemistry
2 answers:
hammer [34]3 years ago
7 0

Answer: C

Explanation:

I think the answers correct because, A says that the food chain will stay the same, but thats not true because, the frog ate most of the insects. And B says the population will decrease, but that doesn't make sense because a animal that eats grasshoppers would no longer be there. And C says that the population will increase and i think that correct because, one animal that does eat the grasshopper wouldn't be there so the grasshoppers will increase in population. And D says the population of snacks will increase but that was one of the foods the snacks ate so they would decrease because they lost one species to eat off of.

Hope This Helps!

sergey [27]3 years ago
3 0
Grasshopper population will increase since they have less predators
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 Best Answer:  <span>(a) 
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Describe the difference between a ball-and-stick model and a space-filling model of a compound.
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4 0
4 years ago
Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la
tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]

2. Derive the rate law

Assume k₋₁ ≫ k₂.  

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.  

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

5 0
4 years ago
What is the difference between a mixture and compound?
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