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3 years ago

How many electrons will a single atom of sulfur with no charge and no bonds have in its valence shell?

Physics

2 answers:

Svet_ta [14]3 years ago

6 0

Answer:

There are 6 electrons in the outermost shell.

Explanation:

Sulphur is a non-mettalic element which is in the period 3 and group .6on the periodic table. It has an atomic number of 16 and a Mass number of 32. Atomic number tells you the number of electrons in an electrically neutral atom. It has the electronic configuration of 1s2 2s2 2p6 3s2 3p4.

The orbitals have a formula 2n^2 where n = 0, 1, 2, 3 etc.

In the shells, n = 1 so there are 2 electrons. For n = 2, 2*(2)^2 = 8 electrons. So, 16 - (8 + 2) = 6 electrons in the 3 shell (outermost shell)

Therefore from the electronic confriguration above, there are 6 electrons in the outermost shell.

ELEN [110]3 years ago

6 0

Answer:

Explanation:

Sulfur is an element that can be found on Group 6 and Period 3 on the periodic table with the Atomic Mass 32 and Atomic Number 16.

In an electronic configuration, the maximum number of electrons in a given shell can be obtained using the formula 2n², where n represent the number of valence shell.

Therefore in the first shell when n=1;

the maximum number of electrons it can contain is 2 × 1² = 2

For the second shell, when n= 2

the maximum number of electrons it can contain is 2 × 2² = 2 × 4 = 8

For the third shell, when n= 3

the maximum number of electrons it can contain is 2 × 3² = 2 × 9 = 18

.......

Now, in a single atom of sulphur with no charge and no bonds, the Atomic Number is said to be 16

We can therefore determine the number of the electrons in the outermost valence shell to be 6 because, the first shell takes 2 electrons, the second shell takes 8 electrons and the third shell takes 6. All together making 16 electrons.

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A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

6 0

3 years ago

Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30

barxatty [35]

For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles

5 0

2 years ago

1. Which of the following is not an acceptable Physical Education activity?

Dmitrij [34]

The answer is A

Explanation:
Vacuuming doesn’t involve a lot of physical movements.

3 0

2 years ago

Read 2 more answers

1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant

Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction $(v_j)$ = 550 mph

Velocity of jet stream from west to east direction $(v_s)=80\ mph$

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

$\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph$

Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0

2 years ago

A small airplane takes on 302 l of fuel. If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?

ella [17]

To calculate the mass of the fuel, we use the formula

$m = V \times \rho$

Here, m is the mass of fuel, V is the volume of the fuel and its value is $V =302 \ L = 302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL$ and $\rho$ is the density and its value of 0.821 g/mL.

Substituting these values in above relation, we get

$m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg$

Thus, the mass of the fuel 247 .94 kg.

8 0

2 years ago