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Zarrin [17]
3 years ago
5

A scientist conducted an experiment and discovered that certain plants grow faster when given a particular amount of fertilizer.

Another scientist conducted the same experiment and got similar results. Which concept does this best illustrate?
Chemistry
2 answers:
Charra [1.4K]3 years ago
7 0
The concept that this best illustrates is replication. 
erica [24]3 years ago
4 0

Answer:

The concept of repetition is best illustrated in the given question.

Explanation:

Let us consider that a  scientist named A conducted an experiment in his or her laboratory. He discovered a new fertilizer, and when the small amount of that when added to the plant it grows very briskly. Similarly, when an another scientist named B performed the similar experiment in the lab and obtained identical outcomes, which means that scientist B is repeating the same experiment that was done by the scientists A. Thus, it explains the concept of repetition.

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It is called a physical change. It is reversible and doesn't affect the chemical make-up.
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WHEN IN FRICTION not useful
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Friction is not useful in Machinery sometimes, Since it leads to the wear and tear of machine parts.
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11. To a 10.0 ml fermentation culture tube, you add 1.0 ml of 10% glucose (MW 180), 0.5 ml of 10% sucrose (MW 342.30), and 8.5 m
trasher [3.6K]

Answer:

kjk

Explanation:

6 0
3 years ago
Predict the formulas of the products and classify the following reaction:
Zina [86]

Answer:

C18H38(s) + O2(g) → CO₂(g) + H₂O(g).

Explanation:

C₁₈H₃₈ is called Octadecane and it is an Alkane as the molecular formula follows the general formula of Alkanes which is CnH2n+2 .

C₁₈H₃₈ reacts with Oxygen and forms Carbon Dioxide (CO₂) and Water (H₂O).

We can say that on oxidation Octadecane gives Carbon Dioxide (CO₂) and Water (H₂O) because on passing Oxygen over Octadecane the above products are formed.

C18H38(s) + O2(g) → CO₂(g) + H₂O(g).

3 0
3 years ago
For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con
Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

A\rightarrow B

Expt    [A] M        [B] M         Rate [M/s]

1          3.40         4.16           1.82*10^-4

2         4.59         4.16           3.32*10^-4

3.        3.40          5.46          1.82*10^-4

Rate = k[A]^{x}[B]^{y}

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2

To find y:

Divide rate of expt 3 by expt 1

\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0

Therefore: x = 2, y = 0

Rate = k[A]^{2}[B]^{0}

To find k

Use rate for expt 1:

k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1

6 0
3 years ago
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