What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?

It must lose two neutrons and become a different isotope.
It must gain two neutrons and become a different isotope.
It must lose two electrons and become an ion.
It must gain two electrons and become an ion.

Answers

Answer 1

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

Answer 2

Answer:

the third option

Explanation:


Related Questions

How does the abundance of hydrogen and helium support the Big Bang Theory?

Answers

It is the proportion predicted to be present in the early universe.

The hydrogen and helium abundance helps us to model the expansion rate of the early universe.

In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.

In big Bang model, the universe is mostly light or protons.

This abundance of hydrogen and helium is consistent with this big bang model. The process of forming this hydrogen and helium is often called big bang nucleosynthesis.

The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.

Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all  it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.

Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.

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Romeo is a 68 kg astronaut. Juliet is a beautiful cosmonaut who is standing on the balcony of a 4.58 x 10^5kg space station that is at rest and out of gas. Romeo is floating 25 meters away from the space station’s center of mass, how strong is the force between Romeo and Juliet?

Answers

Answer:

F =  3.32 x 10⁻⁶ N

Explanation:

The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:

F = Gm₁m₂/r²

where,

F = Force between Romeo and Juliet = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of Romeo = 68 kg

m₂ = mass of space station = 4.58 x 10⁵ kg

r = distance = 25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²

F =  3.32 x 10⁻⁶ N

Two identical conducting spheres each have a charge of 2 C. They have a radius of 0.1 m and are separated by 0.5 meters. If you were to increase the radius of these spheres to 0.2 meters, the electrostatic force between them would

Answers

Answer:

The electrostatic force will remain the same

Explanation:

From the question we are told that  

     The charge on the each conducting sphere is [tex]q = 2 \ C[/tex]

      The radius each sphere is [tex]r_1 = r_2 = r = 0. 1\ m[/tex]

     

Generally electrostatic force between the sphere is mathematically represented as

             [tex]F = \frac{k * q_1 q_2 }{d^2}[/tex]

Here  k is the coulomb constant ,

         d  is the distance between the two sphere which is measured from one center of the sphere to the other center of the sphere  

Now from the question we are told that the radius of the spheres is doubled (i.e  from 0.10 m  to 0.2 m ) , this will not affect the distance between the sphere because their center are still in the same position

and given there is no change in the distance between the spheres , the electrostatic force will remain the same

Our school needs to offer healthier options in the lunchroom. Elever High School has recently updated its cafeteria menu to include whole wheat pasta and breads, a fresh salad bar, and other healthy menu items. Students there claim that they have more energy and focus throughout their school day. Let's encourage healthier menus in our lunchroom!

What type of evidence does the writer of this passage use to support her claim?


statistics
statistics

an expert's opinion
an expert's opinion

examples
examples

the writer's opinion

Answers

Answer:

b

Explanation:

A shell traveling with speed, v0 , exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0 . What is the velocity of the other fragment? Hint: Velocity has both magnitude and direction.

Answers

Answer:

yeah

Explanation:

yeah yeah yeah yeah

A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?

Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198

Answers

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

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In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to

Answers

Answer:

Fall

Explanation:

HELP ASAP !!!!!!!!!!!!!!

Answers

The Big Bang theory supports that belief. If everything exploded from one point, then it is continuing to expand outward now

Answer:

big bang theory

Explanation:

Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.

HELP ASAP !!! !!!!!!!

Answers

the second option - B

Answer:

they are cooler than the rest if the sun

Fast and safe heart rate for workouts is called muscular strength? True or false

Answers

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

PLS HURRY!!!! 15 PTS!!! The pictures represent three different states of matter.
Which order of pictures places molecules with the most
amount of motion to the least amount of motion?
O X Y Z
O ZYX
O Y Z
O Y - X - Z

Answers

Xyz I think!
I’m so sorry if I’m wrong!!!

Please help me with this question guys.

Answers

Answer:

The average speed is 22.2 km/h

Explanation:

Average Speed

Given an object travels a total distance d and took a total time t, then the average speed is:

[tex]\displaystyle \bar v=\frac{d}{t}[/tex]

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]

Then he drives d2=7 km at v2=43 km/h taking a time of:

[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]

The average speed is 22.2 km/h

Catching a wave, a 77 kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?

Answers

Answer:

Explanation:

The total work done by the wave is expressed as;

Workdone = Potential energy + Kinetic energy

Workdone = mgh + 1/2mv²

m is the mass = 77kg

g is the acceleration due to gravity = 9.8m/s²

v is the velocity = 8.2m/s

h is the height = 1.65m

Substitute into the formula;

Workdone = 77(9.8)(1.65) + 1/2(77)8.2²

Workdone = 1245.09 + 2588.74

Workdone = 3833.83Joules

Hence the amount of non conservative work done on the sofa is 3833.83Joules

The non-conservation work done will be "3833.83 Joules".

Given:

Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kg

We know,

→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]

or,

                     [tex]= mgh +\frac{1}{2} mv^2[/tex]

By putting the values,

                     [tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]

                     [tex]= 1245.09+2588.74[/tex]

                     [tex]= 3833.83 \ Joules[/tex]

Thus the above approach is right.

Learn more about work done here:

https://brainly.com/question/24230840

An arrow in a bow has 357 J of elastic potential energy How much Winette enere
Will the arrow have after it has been shot assuming there is no sir restoran

Answers

Answer:

357 J

Explanation:

The elastic potential energy of arrow in the stretched bow is 357 J.

The kinetic energy of the arrow after it has been shot is given by half of the product of the arrow's mass and velocity of the arrow.

Here there are no other forms of energy at play here. Only potential and kinetic energy.

As we know that in any system the energy is conserved accordingly the elastic potential energy of the arrow will be equal to the kinetic energy of the bow after it is released i.e., 357 J.

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