All categories

3 years ago

PLEAS HELP.

1 answer:

8 0

<span>A: put an atom on a poster in the exhibit
   Good luck. The poster itself is made of trillions of trillions of trillions
   of atoms. You could not see the extra one any easier than you could
   see the ones that are already there, and even if you could, it would be
   lost in the crowd.

B: use a life size drawing of an atom
   Good luck. Nobody has ever seen an atom. Atoms are too small
   to see. That's a big part of the reason that nobody knew they exist
   until less than 200 years ago.

D: set up a microscope so that visitors can view atoms
   Good luck. Atoms are way too small to see with a microscope.

</span><span><span>C: Display a large three dimensional model of an atom.
  </span> </span>Finally ! A suggestion that makes sense.
   If something is too big or too small to see, show a model of it
   that's just the right size to see.

You might be interested in

Whats the term for climate near equator

storchak [24]

The closer to the equator, the hotter the climate will be.

6 0

3 years ago

Read 2 more answers

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

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

5 0

3 years ago

The plane of a rectangular coil, 3.7 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

gizmo_the_mogwai [7]

Answer:3.77 T/s

Explanation:

Given

Area of cross-section $=3.7\times 3.7 cm^2=13.69 cm^2$

N=no of turns=61

Resistance= $R=8 \Omega$

current (i)=0.04 A

emf induced $=i\times R=0.04\times 8=0.32 V$

emf induced $e=NA\frac{\mathrm{d} B}{\mathrm{d} t}$

$0.32=62\times 13.69\times 10^{-4}\times \frac{\mathrm{d} B}{\mathrm{d} t}$

$\frac{\mathrm{d} B}{\mathrm{d} t}=\frac{0.32\times 10^4}{848.78}=3.77 T/s$

8 0

3 years ago

A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has

LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

<h3>What is the distance?</h3>

The separation of one thing from another in space; the distance or separation in space between two objects, points, lines, etc.; remoteness. The distance of seven miles cannot be accomplished in one hour of walking.
Learn how to use the Pythagorean theorem to get the separation between two points using the distance formula. The Pythagorean theorem can be rewritten as d==(((x 2-x 1)2+(y 2-y 1)2)
The distance between any two places is the length of the line segment separating them. By measuring the length of the line segment that connects the two points in coordinate geometry, the distance between them may be calculated.

$W_{total} =$ ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1* $2.3^{2}$