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Gnom [1K]
3 years ago
5

the results of Rutherford’s gold foil experiment disproved the model of the atom that his mentor and colleague had proposed. How

does this example represent part of the necessary process for developing a scientific theory?
Chemistry
1 answer:
nignag [31]3 years ago
5 0
During the development of a scientific theory, numerous scientific experiments are carried out through different approaches. A theory holds true as long as no experiment disproves a theory. 
This experiment showed that when testing a theory with a new approach, it is possible that it may be disproven, meaning scientists should not settle for any explanation for a natural phenomenon that is not completely explained; instead, they should continue to investigate and hypothesize to improve their understanding of the phenomenon.
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Explain how the copper could be in the lake sample near the picnic area but not have been detected by this test.
Masja [62]

Answer:

May be the instrument is incorrect or may be error in it.

Explanation:

The copper have not been detected by this test because the test may be not for the detection of copper, may be it is used for identification of another minerals. If there is copper in the lake sample but can't be detected in the test so it means that the instrument which is used for detection is not the right one  or having error in that instrument. Every mineral has a specific type of instrument that detect its presence, if we use incorrect instrument for the mineral then we can't detect the presence of that specific mineral.

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Explanation:

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D. Is Oxidized Hope I helped.
4 0
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Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M
GarryVolchara [31]

Answer:

Explanation:

CH₃Br+NaOH⟶CH₃OH+NaBr

It is a single step bimolecular reaction so  order of reaction is 2 , one for CH₃Br and one for NaOH .

rate of reaction = k x [CH₃Br] [ NaOH]

.008 = k x .12 x .12

k = .55555

when concentration of CH₃Br is doubled

rate of reaction = .555555 x [.24] [ .12 ]

= .016 M/s

when concentration of NaOH  is halved

rate of reaction = .555555 x [.12] [ .06 ]

= .004  M/s

when concentration of both CH₃Br and Na OH is made 5 times

rate of reaction = .555555 x .6 x .6

= 0.2 M/s

8 0
3 years ago
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