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2 years ago

Where are the most reactive nonmetal elements found on the periodic table?

Physics

2 answers:

Sindrei [870]2 years ago

8 0

Answer;

At the top of group 17

Explanation;

Reactivity in chemistry deals with whether something will react with another substance. The reactivity of an element can be predicted based on its location on the periodic table.

The most reactive nonmetal is fluorine. Fluorine is a halogen, which is Group 17 on the periodic table, and the halogens are the most reactive nonmetals. This is because they all have one empty space in their valence electron shells. Of the halogens, fluorine's empty space is closest to the positively charged nucleus. This means it pulls harder on another atom's electrons to fill that empty space.

Llana [10]2 years ago

3 0

Answer: Option (a) is the correct answer.

Explanation:

Elements of group 17 are known as halogens whereas elements of group 18 are known as noble gases.

Noble gases have completely filled valence shell. Therefore, they are stable and does not react easily.

Whereas halogens have incompletely filled sub-shells, therefore, in order to gain stability they react readily with another atom.

Halogen atoms are also electronegative in nature. On moving down the group, the electronegativity decreases. Thus, the atom at the top of group 17, that is, fluorine is the most reactive non-metal because fluorine has high electronegativity so, it readily attracts an electron in order to completely fill its sub-shell.

Thus, we can conclude that at the top of group 17 most reactive nonmetal elements are found on the periodic table.

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A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45

vladimir2022 [97]

<span>31.3 m/s Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know. Distance the water balloon travels at velocity v for time t d = vt Total time required for the entire trip is double since the balloon goes up, then goes down t = 2v/a Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2 t = 2v/9.8 100 = vt Substitute 2v/9.8 for t in the 2nd formula 100 = v(2v/9.8) Solve for v. 100 = v(2v/9.8) 100 = 2v^2/9.8 980. = 2v^2 490 = v^2 22.13594 = v So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that 2*22.13594 / 9.8 = 4.51754 So it will take 4.51754 second for the balloon to hit the ground after being launched. 4.51754 * 22.13594 = 100 And during that time it will travel 100 meters horizontally. But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>

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2 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

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2 years ago

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alina1380 [7]

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2 years ago

Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?

Salsk061 [2.6K]

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

NB: The charges are identical.

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2 years ago

A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it, as shown in the figure above.

cestrela7 [59]

Explanation:

Forgive me If I am wrong, it's been a while since I've studied Torque.

The formula for the angular momentum is

momentum= I*w.

We can also write I*W as 1/2MR^2 * W so the extra mass coming from the block of clay would most likely cause the angular momentum to increase from the amount it was before.

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1 year ago