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VARVARA [1.3K]
3 years ago
8

A student will measure and record the growth of two flowering plants every day for 28 days. Plant A will be watered and fertiliz

ed, Plant B will only be watered. The student predicts fertilizer added to the soil leads to taller flowering plants. Scientific conclusions are based on scientific evidence, what evidence would support the student’s prediction? A) Plant A and Plant B both die B) Plant A grows taller than Plant B C) Plant B grows taller than Plant A D) Plant A and Plant B grow to the same height
Chemistry
2 answers:
mestny [16]3 years ago
5 0
The conclusion that would support the students prediction is B) Plant "A" grows taller than Plant "B".
If the student thinks that adding fertilizer to the plant would help it grow then answer B) would make the most sence.
Jlenok [28]3 years ago
4 0

Answer:

The correct answer will be option-B

Explanation:

In the given question, the experiment was performed to test "if adding the fertilizer will increase the growth of plants" in terms of height and measured the height of the plants.

The experimental group (plant A) was provided with the water and fertilizer whereas the control group (plant B) was provided only water and the height of the plant was recorded which showed that experimental group showed more height.

This can be explained as the fertilizer adds the nutrients to the soil which the plants lack and since the plants get enough amount of fertilizer. Thus, Plant A grows taller than plant  B.

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Answer : The initial rate of the reaction is, 1.739\times 10^{-3}s^{-1}

Explanation :

First we have to calculate the rate constant of the reaction.

Expression for rate law for first order kinetics is given by :

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = ?

t = time taken for the process  = 44 s

[A_o] = initial amount or concentration of the reactant  = 0.1108 M

[A] = amount or concentration left time 44 s = 0.0554 M

Now put all the given values in above equation, we get:

k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}

k=0.0157

Now we have to calculate the initial rate of the reaction.

Initial rate = K [A]

At t = 0, [A]=[A_o]

Initial rate = 0.0157 × 0.1108 = 1.739\times 10^{-3}s^{-1}

Therefore, the initial rate of the reaction is, 1.739\times 10^{-3}s^{-1}

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