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alexandr1967 [171]
3 years ago
11

Which best explains why elements with a completely full outer shell of electrons do not usually bond with other elements to form

compounds?
A.
A full outer shell of electrons indicates a stable state, making the formation of compounds unnecessary. The full outer shell was obtained by releasing electrons instead of forming a compound.

B.
These elements have obtained electrons from other elements without bonding to create a full outer shell of electrons, and a stable state.

C.
Elements with a completely full outer shell of electrons are in their most stable state. Therefore, achieving stability through the formation of compounds is not necessary.

D.
Because electrons orbit some atoms in three dimensional space, they move to different orbitals to stabilize the atom, making the formation of compounds unnecessary.
explain.
Physics
1 answer:
Alla [95]3 years ago
8 0

The best answer is C.

The stability of atoms depends on whether or not their outermost shell is filled with electrons. If the outer shell is filled with electrons, the atom is stable and therefore they do not need to react  with other elements to become stable.

On the other hand, atoms with unfilled outer shells are unstable, and will usually form chemical bonds with other atoms to achieve stability. To achieve stability, atoms will form two types of chemical  bonds called ionic bonds and covalent bonds.





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Two waves collide and the temporary combined waves that results is smaller that the original wave.
zimovet [89]
A.

Destructive interference is when two waves cancel each other out or when the crest of one wave passes through the trough of another wave.
5 0
3 years ago
Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown
jonny [76]

Answer:

Precipitation is the formation of a solid from a solution. It is necessary to centrifuge the precipitate to exert sufficient forces of gravity to bring the solid particles in the solution to come together and settle

Explanation:

When you centrifuge precipitate it enables the nucleation to form.

Centrifuging the precipitate helps in determining whether a certain element is present in a solution or not.

5 0
3 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
Which of these tools is used to measure temperature?
Feliz [49]

Answer:

D a thermometer

Explanation: It measures and track Celcius and Feirinheit.

3 0
3 years ago
Read 2 more answers
A 1.40-kg ball tied to a string fixed to the ceiling is pulled to one side by a force F→ . where L = 1.40 kg. What is the tensio
riadik2000 [5.3K]

Answer:

T=13.72N

Explanation:

The tension before the ball is released have no angle is in rest at the same axis of the weight so:

∑F=0

Using Newton law in this case the ball is tied so tension before become to swing is

∑F=FN-T=0

T=F_{N}

T=m*g

T=1.40Kg*9.8\frac{m}{s^2}

T=13.72N

8 0
3 years ago
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