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Zolol [24]
3 years ago
10

Which statement correctly compares the motion of atoms in liquid mercury and solid copper? Copper atoms in copper move while sta

ying close together, whereas mercury atoms vibrate around fixed positions. Copper atoms move while staying close together, whereas mercury atoms are independent of one another. Copper atoms vibrate around fixed positions, whereas mercury atoms move around while staying close together.
Chemistry
2 answers:
lapo4ka [179]3 years ago
5 0

Answer:

Copper atoms vibrate around fixed positions, whereas mercury atoms move around while staying close together.

Explanation:

rewona [7]3 years ago
4 0

Answer:

  • <u><em>Copper atoms vibrate around fixed positions, whereas mercury atoms move around while staying close together. </em></u>

Explanation:

The <em>solid </em>state of matter is characterized by rigid crystalline structures, in which <em>atoms </em>can vibrate around a fixed position. That is why <em>solids</em> have definite shape and volume, and cannot flow. This is described in the last statement when it states <em>"Copper atoms vibrate around fixed positions"</em>

The <em>liquid </em>state of matter is characterized by the <em>atoms </em>being more separated than in<em> solids.</em> While the <em>atoms </em>in the liquid are not completely free to move, they can slide past the neighboring atoms, which gives them the property to flow. Therefore, the liquids, although they have a defined volume, take the form of the container. This is what the last statement describes as "mercury atoms move while remaining together"

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What type of bonds are shown in this diagram?
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Answer:

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A: metallic bonds and covalent bonds

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Explanation:

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3 0
1 year ago
Read 2 more answers
Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2
IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at 75^{o}F and 14.6 psia.

\varepsilon = 0.00015 ft,     Flow rate, (Q) = 48000 ft^{3}/m

(a)  Formula to calculate hydraulic radius (r_{H}) is as follows.

              r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}

                          = \frac{2 \times 1}{2(1) + 2(2)}

                          = \frac{1}{3} ft

Formula for equivalent diameter is as follows.

                     D_{eq} = 4 \times r_{H}

                                    = 4 \times \frac{1}{3} ft  

                                    = \frac{4}{3} ft

(b)    Formula for velocity floe is as follows.

                         Q = VA

                     V = \frac{Q}{A}

                        = \frac{48000}{2 \times 1} ft/min

                        = 24000 ft/min

(c)   Formula to calculate Reynold's number is as follows.

         R_{e} = \frac{D \times V \times \rho}{\mu}

                   = \frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}  (as \rho = 0.0744 lb/ft^{3} and \mu = 0.0443 lb/ft. hr)

                   = 53742.66 hr/min  

As 1 hr = 10 min. So, 53742.66 hr/min \times \frac{60 min}{1 hr}

                            = 3224559.6

(d)   Formula to calculate pressure drop (\Delta P) is as follows.

              \frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}

Putting the given values into the above formula as follows.

               \frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}

                      = \frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}

                      = 6.238 lb/ft^{2}

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Hope this helped! Please rate 5 starts and hit thanks if it did :)
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