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Brums [2.3K]
3 years ago
13

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

Physics
2 answers:
Sindrei [870]3 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]3 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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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless
drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

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sertanlavr [38]

Answer:

There is no mechanical advantage

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The mechanical advantage is possible only when the force needed to lift a load is lesser than the weight of the load.

For example, is we have a mechanical advantage of 2, the force needed to lift will be 1/2 of the weight of the load, and if we have a mechanical advantage of 4, the force needed will be 1/4 of the weight of the load.

In the attached image there are clear examples of mechanical advantage with pulleys.

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