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Otrada [13]
2 years ago
14

Which statement correctly compares the valence electrons for atoms of francium (Fr) and barium (Ba)?

Physics
2 answers:
Sauron [17]2 years ago
8 0

Answer:

Francium has fewer valence electrons, but they are in a higher energy level

34kurt2 years ago
5 0

From the Periodic Table...

Francium is a group 1 Element and Barium is a Group 2 Element

This statement is a giveaway as it already tells you that Barium has more valence electrons than Francium since it has 2 electrons in its Outer shell.

This Eliminates Options A and B

Francium has more electrons than Barium

So It should be at a higher energy level.

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Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of
Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

V = 21\% * 0.66L

V = 0.1386L

Our values are given as

P = 1atm

T=293K R = 0.083145kJ*mol^{-1}K^{-1}

Using the equation to find n, we have:

PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

7 0
3 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
As the distance between the sun and earth decreases, the speed of the planet
mixas84 [53]

Answer:

Explanation:

Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.

5 0
3 years ago
Which events took place during the Copernican revolution, when most people started to believe in a heliocentric model of the sol
balu736 [363]
<h2>2) Copernicus rediscovered Aristarchus’s heliocentric model.</h2>

Before Copernican Revolution, people did believe in the ptolemain model that establishes the description of the Universe with the earth at the center having sun, moon, starts and planets all orbited earth. On the other hand, the heliocentric model establishes the sun at the center of the solar system and this starts with the publication of Nicolas Copernicus named <em>De revolutionibus orbium coelestium.</em>

<h2>5) Newton’s theories of gravity increased understanding of the movement of planets.</h2>

The revolution ended with Isaac Newton's work over a century later. As you well know, Newton was both a physicist and mathematician, better known for his prodigal work called <em>Philosophiæ Naturalis Principia Mathematica. </em>In this revolution, he is known for his laws of motion and universal gravitation increasing understanding of the movement of planets.

7 0
2 years ago
Read 2 more answers
Why was the Roman empire able to prosper
gayaneshka [121]

Answer:

The mild climate enabled Romans to grow wheat, grapes, and olives. This abundance o food supported the people and allowed Rome to prosper. While the climate made year-long agriculture possible, Rome also had the advantage to be near water. The Tiber River helped the agricultural system to prosper

Explanation:

3 0
3 years ago
Read 2 more answers
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