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NARA [144]
3 years ago
10

A scientist wants to perform a controlled experiment to test the effect of rust on chemically treated iron. which action would b

e most important to perform in a controlled experiment? design an experiment that is based on previously published results. expose the treated iron to the corrosive effects of the air. test a large piece of treated iron and a small piece of untreated iron. test a piece of treated iron and a piece of untreated iron under the same conditions.
Chemistry
2 answers:
olga2289 [7]3 years ago
9 0
The answer is to test a piece of untreated and treated under same conditions!
Ilia_Sergeevich [38]3 years ago
8 0

Answer: Option (d) is the correct answer.

Explanation:

A test or experiment in which only one variable is changed at a time in order to verify the result is known as a controlled experiment.

That is only independent variable need to be adjusted with respect to the dependent variable.

Thus, we can conclude that the action to test a piece of treated iron and a piece of untreated iron under the same conditions would be most important to perform in a controlled experiment in the given situation.


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Which of the following represents the velocity of a light wave?
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3 0
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Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos
SCORPION-xisa [38]

Explanation:

More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a)   Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, PF_{3} is the most ionic compound and PBr_{3} is the least ionic compound.

(b)   Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, BF_{3} is the most ionic compound and NF_{3} is the least ionic compound.

(c)   Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, TeF_{4} is the most ionic compound and BrF_{3} is the least ionic compound.

6 0
3 years ago
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