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4 years ago

When very electronegative atoms, like oxygen, bond to atoms with lower electronegativity, like lithium, what's the result?

Physics

1 answer:

umka2103 [35]4 years ago

3 0

(A) The oxygen atom becomes strongly negative.

Explanation:

When a very electronegative atom like oxygen bonds to an atom with a lower electronegativity, like lithium, the oxygen atom becomes strongly negative and the lithium becomes strongly positive.

The attraction between these oppositely charged ions results in the formation ionic bond.

An ionic bond is an interatomic bond in which a highly electronegative atom is bonded to less electronegative one.
The electronegative nature of the oxgyen allows is to gain the electrons lost by weakly electronegative lthium.
Most metals have low electronegativity.
They then both will form ions.
The attraction between the oppositely charged ions results in an ionic/electrovalent bonding.

learn more:

Ionic bond brainly.com/question/6071838

#learnwithBrainly

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A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3

Ipatiy [6.2K]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

$\Delta P \propto \rho \cdot \Delta h$

$\Delta P = k \cdot \rho \cdot \Delta h$

Where:

$\Delta P$ - Manometric pressure difference, measured in kilopascals.

$\rho$ - Fluid density, measured in kilograms per cubic meter.

$\Delta h$ - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

$\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}$

$\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}$

Where:

$\Delta h_{air}$ - Height difference of the air column, measured in meters.

$\Delta h_{Hg}$ - Height difference of the mercury column, measured in meters.

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\rho_{Hg}$ - Density of mercury, measured in kilograms per cubic meter.

If $\Delta h_{Hg} = 0.015\,m$ , $\rho_{air} = 1.29\,\frac{kg}{m^{3}}$ and $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , the height difference of the air column is:

$\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)$

$\Delta h_{air} = 158.140\,m$

The height of the building is 158.140 meters.

5 0

3 years ago

Read 2 more answers

A merry-go-round accelerates from rest to 0.75 rad/s in 33 s.

Aleks04 [339]

Answer:

1309.1 Nm

Explanation:

Torque is given as a product of Moment of innertia and acceleration hence

T=Ia where T is torque and a is acceleration

To get acceleration, it is rate of change of speed per unit time hence $a=\frac {v_f-v_i}{t}$ where v and t represent velocity and time respectively while subscripts f and i represent final and initial respectively. Also, I is given by $0.5mr^{2}$ where m js mass and r is radius hence the net torque can now be written as

$T=0.5mr^{2}\times \frac {v_f-v_i}{t}$

By substituting the given figures then

$T=0.5\times 3.2\times 10^{4}\times 6^{2}\times \frac {0.75-0}{33}=1309.0909090867 Nm\approx 1309.1 Nm$

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3 years ago

A diver who is 10.0 m underwater experiences a pressure of 202 kPa. If the diver's surface area is 1.50 m.sq, with how much tota

Deffense [45]

The pressure on the diver is given by:

$p=\frac{F}{A}$

where F is the force exerted by the water on the diver, while A is the area of the diver.

In this problem, we know the pressure: $p=202 kPa=2.02 \cdot 10^5 Pa$ , and the area of the diver, $A=1.50 m^2$ , therefore we can rearrange the initial equation to find the force on the diver:

$F=pA=(2.02 \cdot 10^5 Pa)(1.50 m^2)=3.03 \cdot 10^5 N$

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4 years ago

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Coulomb is a very large unit for practical use. Justify your answer if 10^10 electrons are transferred from a body/second

zlopas [31]

Given:

10^10 electrons per second

To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:

From literature,

1 Coulomb is equivalent to 6.242×10^18 electrons.

So,

= 10^10 electrons * (1 coulomb/6.242×10^18 electrons) / second
= 1.602 x 10^-9 coulumbs

This value is too small to be used in an actual setting.

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3 years ago

List the planet name and position from the Sun for each one (1, 2, 3)

IgorLugansk [536]

Answer:

The planets in order from the sun are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and finally the dwarf planet Pluto. Most people have at least heard about our solar system and the planets in it.

Explanation:

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