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elena-14-01-66 [18.8K]
3 years ago
9

What is the relationship between the valence electrons of an atom and the chemical bonds the atom can form?​

Physics
1 answer:
stellarik [79]3 years ago
8 0

Answer:

Valence electrons are outer shell electrons with an atom and can participate in the formation of chemical bonds. In single covalent bonds, typically both atoms in the bond contribute one valence electron in order to form a shared pair. The ground state of an atom is the lowest energy state of the atom.

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Exposure to the Sun's harmful infrared radiation should be kept to a minimum.
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Of course! If it's harmful, then your exposure to it should be kept
to a minimum.  That's a no-brainer.  But the sun's infrared radiation
is generally less harmful than its ultraviolet radiation is.

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3 years ago
For the following questions, imagine the book drops in free fall and falls a distance d. For the system of the book and earth, t
alukav5142 [94]

Answer:

C) is zero

Explanation:

According to the law of energy conservation, the total mechanical energy of the object is conserved. A book falling a distance d would have a change in potential energy, resulting in the same change in kinetic energy. But the total mechanical energy must be the same. So there's 0 change in total energy of the system.

8 0
3 years ago
If you run 10 km east and then turn around and run 7 km west, how much total distance have you travelled? A. 17 km B. 70 km C. 3
Tom [10]

It's A 17 km because the total km walked is 17.

4 0
3 years ago
Read 2 more answers
A cheetah starts from rest and accelerates after a gazelle at a rate of 6.5 meters per second2for 3.0 seconds. Calculate the che
Pepsi [2]

Answer:

the speed of the cheetah at the end of the 3 seconds is: 19.5 m/s

Explanation:

Let's use the equation that relates speed with acceleration:

vf = vi + a * t

where vf stands for final velocity, vi stands for initial velocity, a for acceleration, and t for the time acceleration is applied. Then, in our case we have:

vf = 0 + 6.5 (3)

vf = 19.5 m/s

6 0
3 years ago
A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic
xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2

 =2.5m

on the axis of x at x = 2.5

r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}

r = 2.5m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T

(Along z direction)

B)r = 5.00m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2

 =5.00m

on the axis of x at x = 5.0

r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}

r = 5.00m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T

(Along x direction)

7 0
3 years ago
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