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Lisa [10]
3 years ago
14

TIMED TEST PLEASE HURRY!!! The development of a new experimental method is most likely to change a theory if it makes it possibl

e to
analyze samples when they are frozen.

perform the experiment in a different lab.

obtain and analyze results more quickly.

study a larger sample size than before.
Chemistry
2 answers:
Vikki [24]3 years ago
6 0

Answer: is the D

Explanation:

DerKrebs [107]3 years ago
5 0

Answer:

study a larger sample size than before.

Explanation:

Science is evolving day by day and every new experiment is contradict the old experiment, giving new theories.

The advancement in the technology modifying the new experimental methods and it can change a theory if study a larger sample size than before. It is so because new experiment verify and validates larger sample size and evidences can contradict the old theory. Large sample size means more experiments and more validations.

Hence, the correct answer is "study a larger sample size than before."

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Answer:

Only Fe^{3+}+Mg gives spontaneous reaction.

Explanation:

A redox reaction will be spontaneous if standard reduction potential (E^{0}) of the reaction is positive. Because it leads to negative standard gibbs free energy change (\Delta G^{0}), which is a thermodynamic condition for spontaneity of a reaction.

E^{0}=E^{0}(reduction)-E^{0}(oxidation)

Where E^{0}(reduction) and E^{0}(oxidation) represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation:         Mg-2e^{-}\rightarrow Mg^{2+} ;  E_{Mg^{2+}\mid Mg}^{0}=-2.38V

Reduction:         Fe^{3+}+3e^{-}\rightarrow Fe ; E_{Fe^{3+}\mid Fe}^{0}=-0.04V

So, E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V

Hence this pair will give spontaneous reaction.

(2) Similarly as above, E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V

Hence this pair will give non-spontaneous reaction.

(3) Similarly as above, E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V

Hence this pair will give non-spontaneous reaction.

(4)  Similarly as above, E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V

Hence this pair will give non-spontaneous reaction.

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