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kati45 [8]
3 years ago
8

Good experimental design includes a hypothesis that is

Physics
2 answers:
scZoUnD [109]3 years ago
7 0

Answer: testable and observable.

Explanation:

Experimental design includes the concepts which are directly related to the plan which is required to be followed so as to deduce a reasonable outcome. It must include a hypothesis which should be testable and observable on the basis of application of suitable scientific methodology. A hypothesis is presumed statement related to a particular natural phenomena or event which is required to be proved by implementation of scientific experimentation procedure and direct observation.

Ratling [72]3 years ago
3 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Good experimental design includes a hypothesis that is <span>testable and observable.</span>
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71.6 <span>°F</span>
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A child on a high dive has a mass of 40 kilograms. If the high dive is 10 meters in the air, what is the potential energy? GPE=m
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Answer:

Ep = 3924 [J]

Explanation:

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xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi
slega [8]

Answer:

Approximately \rm 90\; kJ.

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (E^{\circ}(\text{cell})) is equal to

E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode}).

There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

\rm Cu^{2+} + Pb \to Cu + Pb^{2+}.

Its cell potential would be equal to 0.339 - (-0.130) = \rm 0.469\; V.

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell}),

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  • n is the number moles of electrons transferred for each mole of the reaction. In this case the value of n is 2 as in the half-reactions.
  • F is Faraday's Constant (approximately 96485.33212\; \rm C \cdot mol^{-1}.)

\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}.

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