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Physics

1 answer:

nignag [31]3 years ago

5 0

Answer:

measures of central tendency are numbers that describe what an average or typical within a distribution of data there are three main measures of central tendency mean medium and mold well they are all measuring essential tendency each is calculated differently and measures something different from others

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Provided following are four different ranges of stellar masses. Rank the stellar mass ranges based on how many stars in each ran

elena-s [515]

Highest to lowest number:

-less than 1 solar mass

-between 1 and 10 solar masses

-between 10 and 30 solar masses

-between 30 and 60 solar masses

<h3>What is Stellar masses ?</h3>

Stellar mass is a phrase that is used by astronomers to describe the mass of a star.

It is usually enumerated in terms of the Sun's mass as a proportion of a solar mass ( M ☉). Hence, the bright star Sirius has around 2.02 M ☉.

Stellar masses are not fixed, although they change for single stars only on long periods.

Learn more about Stellar masses here:

brainly.com/question/1128503

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3 0

2 years ago

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:

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

where

$n$ 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 $96485.33212\; \rm C \cdot mol^{-1}$ .)

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