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3 years ago

when an element tends to lose its valence electrons in chemical reactions , does it behave more like a metal or nonmetal

Physics

1 answer:

Juli2301 [7.4K]3 years ago

4 0

It behaves more like a metal

Explanation:

When an element tends to lose its valence electrons in chemical reactions, they behave more like a metal.

Metals are electropositive.

Electropositivity or metallicity is the a measure of the tendency of atoms of an element to lose electrons.

This is closely related to ionization energy and the electronegativity of the element.

The lower the ionization energy of an element, the more electropositive or metallic the element is .

Metals are usually large size and prefers to be in reactions where they can easily lose their valence electrons.

When most metals lose their valence electrons, they attain stability.

Non-metals are electronegative. They prefer to gain electrons.

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At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at 1.5 atm pressure an

quester [9]

Answer:

$\vartheta = 41.31 GHz$

Given:

Pressure, $P_{O} = 1.5 atm = 1.5\times 10^{5} Pa$

Temperature, T = $- 0.77^{\circ}$ = 273 + (- 0.77) = 272.23 K

Diameter of oxygen molecule, $d_{O} = 8.4\times 10^{- 8} cm = 8.4\times 10^{-10} m$

Speed of sound, v = 330 m/s

$k_{B} = 1.38\times 10^{- 23} J/s$

Solution:

Mean free path of oxygen is given by:

$L_{O} = \frac{k_{B}T}{\sqrt{2}P_{O}\pi d_{O}^{2}}$

Now, substituting the given values in the above formula:

$L_{O} = \frac{1.38\times 10^{- 23}\times 272.23}{\sqrt{2}\times 1.5\times 10^{5}\pi\times (8.4\times 10^{- 10})^{2}}$

$L_{O} = 7.99\times 10^{-8} m$

Now, the frequency, $\vartheta$ is given by:

$\vartheta = \frac{c}{\lambda}$

Since, mean free path = wavelength = $7.99\times 10^{-9} m$

Therefore,

$\vartheta = \frac{v}{L_{O}}$

$\vartheta = \frac{330}{7.99\times 10^{-8}}$

$\vartheta = 4.131\times 10^{10} Hz = 41.31 GHz$

5 0

3 years ago

Reasons why inductors opposes charges passing through it<br>

saul85 [17]

Answer:

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Explanation:

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8 0

2 years ago

For the circuit shown in the figure(figure 1) find the current through each resistor. Express your answers using two significant

Angelina_Jolie [31]

The current flowing in each resistor of the circuit is 4 A.

<h3>Equivalent resistance of the series resistors</h3>

The equivalent resistance of the series circuit is calculated as follows;

6 Ω and 4 Ω are in series = 10 Ω

5 Ω and 10Ω are in series = 15 Ω

<h3>Effective resistance of the circuit</h3>

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \\\\R = \frac{R_1R_2}{R_1 + R_2} \\\\R = \frac{10 \times 15}{10 + 15} \\\\R = 6 \ ohms$

<h3>Current flowing in the circuit</h3>

V = IR

I = V/R

I = 24/6

I = 4 A

Learn more about resistors in parallel here: brainly.com/question/15121871

8 0

2 years ago

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin

tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

cos 40 = v₀ₓ / V₀

v₀ₓ = v₀ cos 40

v₀ₓ = 50.0 cos 40

v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

Y Axis y

$p_{oy}$ = 0

After the break

X axis

$p_{fx}$ = m₂ v₂

Axis y

$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

8 0

3 years ago

A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul

topjm [15]

Answer:

increase in temperature of water is 10° C

Explanation:

Given data

pizza = 500 kcal = 500000 calories

cold water = 50 L

to find out

increase in temperature of water

solution

we know heat formula that is

heat = m × specific heat × Δt

here m is mass = 50000 gram and Δt is change in temperature

and specific heat = 1 cal / gram C

so put here all value and find Δt

500000 = 50000 × 1 × Δt

Δt = 10° C

so increase in temperature of water is 10° C

4 0

3 years ago

Read 2 more answers