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3 years ago

Why does a lone pair of electrons occupy more space around a central atom than a bonding pair of electrons?

Physics

1 answer:

Inessa [10]3 years ago

5 0

Answer:

The lone pair of electrons occupy more space because the electrostatic force becomes weaker.

Explanation:

When there is a bond pair of electrons in the 2 positively charged the atomic nuclei draw the electron density towards them, thereby reducing the bond diameter.

In the case of the lone pair, only 1 nucleus is present, and the enticing electrostatic force becomes weaker and the intensity of the electrons will be increases. Therefore, the lone pair occupies more space than the pair of bonds.

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A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

$y=y_{0}+uSin\theta t-1/2 gt^{2}$

$y=2+20Sin5 t-4.9t^{2}$

As it hits the ground in time t, so put y = 0

$0=2+1.74 t-4.9t^{2}$

$4.9t^{2}-1.74t-2=0$

$t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}$

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

$y'=y_{0}+uSin\theta t'-1/2 gt'^{2}$

$y'=2+20Sin5 t-4.9\times0.35^{2}$

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

3 0

3 years ago

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0

3 years ago

Students are asked to create roller coasters for marbles. Their goal is to design a coaster with the tallest possible hill that

ivann1987 [24]

Answer:

Kinetic Energy.

Explanation:

The movement of a roller coaster is accomplished by the conversion of potential energy to kinetic energy. The roller coaster cars gain potential energy as they are pulled to the top of the first hill. As the cars descend the potential energy is converted to kinetic energy.

3 0

3 years ago

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

kaheart [24]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

$\Delta T$ is the temperature of difference that we want to find

$\Delta x=2.5 cm =0.025 m$ represent the thickness of the material

If we solve $\Delta T$ in absolute value from the equation (1) we got:

$\Delta T =\frac{Q \Delta x}{Ak}$

First we convert 3KW to W and we got:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

And we have everything to replace and we got:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

5 0

3 years ago

You have a 1.55 kg block of silver at room temperature of 20 °C. Silver melts at 962°C. How much heat energy would be needed to

notsponge [240]

Specific heat capacity= heat energy/mass×temperature rise

962°C - 20°C = 942K

Heat energy (Eh) = 239 × 1.55 × 942
Eh= 348963.9J

shc of Ag: 238.6 J/kg-K

m of Ag: 1.55kg

5 0

3 years ago