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3 years ago

What structural formula represents 4 electrons shared between two atoms?

1 answer:

4 0

Sharing of 4 electrons between two atoms results in two double bonds. This can be see in the case of oxygen molecule (O2)

Atomic number of O = 8

Electron configuration of O = 1s²2s²2p⁴

Valence electron configuration: 2s²2p⁴

When 2 O atoms combine they share 4 electrons to form 2 double bonds. In addition, there are two lone pairs on each O atom.

Structural formula: O=O

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Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an

Mumz [18]

Answer:

The hoop

Explanation:

We need to define the moment of inertia of the different objects, that is,

DISK:

$I_{disk} = \frac{1}{2} mR^2$

HOOP:

$I_{hoop} = mR^2$

SOLID SPHERE:

$I_{ss} = \frac{2}{5}mR^2$

HOLLOW SPHERE

$I_{hs} = \frac{2}{3}mR^2$

If we have the same acceleration for a Torque applied, then

$mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2$

$I_{hoop}>I_{hs} >I_{disk}>I_{ss}$

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

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3 years ago

PHYSICS 50 POINTS PLEASE HELP

babymother [125]

Newton’s First Law of Motion - if an object is at rest, it takes un-

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seraphim [82]

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3 years ago

According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

Andrew [12]

Answer:

Part a)

$\rho = 0.55 g/cm^3$

Part b)

$\rho_L = 1.49 g/cm^3$

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

$F_b = \rho_w V_{sub} g$

$F_b = (1 g/cm^3) (30 - 13.5) Ag$

$F_b = 16.5 Ag$

Now we know that the weight of the cylinder is given as

$W = \rho (30 cm)A g$

now we have

$\rho (30 cm) A g = 16.5 A g$

$\rho = 0.55 g/cm^3$

Part b)

When the same cylinder is floating in other liquid then we will have

$F_b = \rho_L (30 - 18.9 )A g$

so we have

$\rho_L (11.1) Ag = 0.55(30) Ag$

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Part c)

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miskamm [114]

Im 99% sure
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