HELP ME PLS!!!!Find the location of beryllium (Be) on the periodic table. What type of ion will
beryllium form?
A. An ion with a -2 charge
B. An ion with a +6 charge
C. An ion with a +2 charge
D. An ion with a -6 charge

Answers

Answer 1

Answer:

The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.

What are cations and anions?

In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.

This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.

Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.

The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.

Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.

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Answer 2

Answer:

Answer:

the answer is c which is a+2 charge

Explanation:

Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.

Related Questions

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An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?
While testing at 30 feet below the surface in Lake Minnetonka, with the sub stopped and in equilibrium, one of the students aboard the sub drops a hammer that goes through the hull of the submarine, and sticks out of the submarine handle first. When this happens, a seal forms immediately around the handle, so that no water enters the sub. What is the new equilibrium position for the sub?

The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force

Answers

Answer:

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K = 50N/m

New length = 32cm

Unknown

Force applied = ?

Solution:

The force applied on a spring can be derived using the expression below;

Force = KE

k is the spring constant

E is the extension

extension = new length - original length

extension = 32cm - 22cm = 10cm

convert the extension from cm to m;

100cm = 1m;

10cm will give 0.1m

So;

Force = 50N/m x 0.1m = 5N

Final answer:

To calculate the force used to stretch the spring, Hooke's Law is utilized, which leads to the conclusion that a force of 5 N was exerted to stretch the spring from its original length of 22 cm to a final length of 32 cm.

Explanation:

The force exerted by a spring is governed by Hooke's Law, which states that the force required to stretch or compress a spring by a certain distance is proportional to that distance. In this case, the spring constant, k, is given as 50 N/m and the spring is stretched from its original length of 22 cm to a final length of 32 cm. This represents a stretch, or displacement, of 10 cm (or 0.1 m when converted to the standard unit).

The force (F) can be calculated using Hooke's law: F = kx, where x is the displacement of the spring. Substituting the given values, the force amounts to F = (50 N/m) * (0.1 m) = 5 N. Therefore, the force used to stretch the spring to its final length of 32 cm is 5 N.

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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

Answers

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 ((v_0)/(v_0-v))$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 ((v_0)/(v_0-v))$

$300 = f_0 ((343)/(343-v))$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 ((v_0)/(v_0-v/2))$

$290 = (343)((v_0)/(343-v/2))$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$((300*343) - 300v)/(290*343-145v) = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

Final answer:

The speed of the train can be determined using the Doppler effect formula.

Explanation:

The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

f' = f((v + v_o)/(v - v_s))

Where:

f' is the observed frequency
f is the source frequency
v is the speed of sound
v_o is the speed of the observer (here it is the train)
v_s is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

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What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

Answers

The mass of a neutron is:
$m=1.67 \cdot 10^(-27)kg$
Since we know its speed, we can calculate the neutron's momentum:
$p=mv=(1.67 \cdot 10^(-27)kg)(1.90 \cdot 10^3 m/s)=3.17 \cdot 10^(-24) kg m/s$

The problem says the photon has the same momentum of the neutron, p. The photon momentum is given by
$p= (h)/(\lambda)$
where h is the Planck constant, and $\lambda$ is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:
$\lambda= (h)/(p)= (6.6 \cdot 10^(-34)Js)/(3.17 \cdot 10^(-24) kg m/s)=2.08 \cdot 10^(-10) m$

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:
$f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(2.08 \cdot 10^(-10) m)= 1.44 \cdot 10^(18) Hz$

Beverage can is thrown upward and then falls back down to the floor. As usual, a y axis extends upward (positive direction). Which of the following best describes the acceleration of the can during its free flight?a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
b) +9.8 m/s^2 throughout
c) -9.8 m/s^2 throughout
d) zero throughout
e) +9.8 m/s^2, then momentarily zero, then -9.8 m/s^2

Answers

a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s

Two identical metal spheres a and b are in contact. both are initially neutral. 1.0×1012 electrons are added to sphere a, then the two spheres are separated. you may want to review ( pages 639 - 641) . part a afterward, what is the charge of sphere a?

Answers

The charge on the sphere A and sphere B after they are separated is $\boxed{ - 80\,{\text{nC}}}$ each.

Further Explanation:

Given:

The number of electrons transferred to sphere $A$ is $1.0 * {10^(12)}$ .

Concept:

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $- 1.6 * {10^( - 19)}\,{\text{C}}$ .

So, the amount of charge carried by the $1.0 * {10^(12)}$ electrons is given as.

$\begin{aligned}Q&= \left( {1.0 * {{10}^(12)}} \right)\left( { - 1.6 * {{10}^( - 19)}} \right)\n&= - 1.6 * {10^( - 7)}\,{\text{C}}\n\end{aligned}$

Since the charge is disturbed equally on the two sphere, so the amount of charge carried by each sphere s half of the total charge.

$\begin{aligned}{Q_A}&= \frac{{ - 1.6 * {{10}^( - 7)}}}{2}\n&= - 8 * {10^( - 8)}\,{\text{C}}\n &= - 8{\text{0}}\,{\text{nC}}\n\end{aligned}$

Thus, the amount of the charge carried by each sphere after separating from each other is $\boxed{ - 80\,{\text{nC}}}$ .

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords: Metal spheres, two identical, in contact, neutral, charged, electrons, charge on electron, charge on metallic sphere, charge of sphere A.

The charge on the sphere A and sphere B after they are separated is $-80\mu C$ each

What is Charge?

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $-1.6* 10^(-19) \ C$