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2 years ago

These diagrams show two atoms of fluorine and an atom of magnesium.

Physics

2 answers:

Morgarella [4.7K]2 years ago

8 0

The correct steps to the following question will be- A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion → Each fluorine atom becomes a -1 ion.

EXPLANATION:

The electronic configuration of fluorine atom is $1s^22s^22p^5$ .

The electronic configuration of magnesium is $1s^22s^22p^63s^2$ .

The valence shell of fluorine needs one electron to satisfy inert gas configuration to be stable. On other hand, the magnesium has two electrons in its valence shell. It needs six electrons more to be stable. As it is a electropositive metal, so it will lose two of its electrons to acquire the nearest inert gas configuration i.e neon.

Hence, magnesium will lose two of its valence electrons and becomes a positively charged cation having +2 charge.

Each of the fluorine atoms will accept one electron from magnesium and becomes negatively charged anion having one unit negative charge.

Now both the cation and anion are attracted by the coulombic force of attraction and forms magnesium chloride.

Bezzdna [24]2 years ago

4 0

the answer is D. A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion → Each fluorine atom becomes a -1 ion

because magnesium only has 2 valence electrons, and it would be easier to lose electrons rather gain.

i just took the test, its right.

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A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 

Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
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2 years ago

Read 2 more answers

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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2 years ago

Suppose you are measuring the height of a small child. What will determine the number if significant digits you record?

slega [8]

The number of significant digits of any measurement is determined by the instrument used for such measurement. For example, in this case, we have the height of a small child being measured. We can use a simple ruler for this, and we see that a ruler has ten divisions for 1 cm. This means that the ruler cannot measure beyond the size of 0.1 cm or 1 mm. Hence, when we report the height of the small child, we report it to one significant digit after the decimal place. As an example, if we measure a child's height to be 90 full cm divisions and 8 smaller divisions, we report it as 90.8 cm but not 90.83 or 90.86 cm.

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2 years ago

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timurjin [86]

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1 year ago

A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

Ulleksa [173]

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$\epsilon = -\frac{\Delta \Phi_B}{\Delta t}$ (1)

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$\Delta \Phi_B$ is the variation of magnetic flux through the coil

$\Delta t = 2.0 ms = 0.002 s$ is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

$\Phi_B = BA$

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while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

$\Phi_f = 0$

so the variation of magnetic flux is

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In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

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2 years ago