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3 years ago

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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A block of mass 0.84 kg is suspended by a string which is wrapped so that it is at a radius of 0.061 m from the center of a pull

lora16 [44]

Answer:

$E_l = 1.713 J$

Explanation:

Given data:

mass of block is $M_b = 0.84 kg$

radius of block = 0.061 m

moment of inertia is $6.20 \times 10^{-3} kg m^2$

D is distance covered by block = 0.65 m

speed of block is 1.705 m/s

From conservation of momentum we have

$M_b g D = \frac{1}{2} M_b v^2 + \frac{1}{2} I \omega^2 + E_{loss}$

$0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times 0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l$

solving for energy loss

$E_l = 1.713 J$

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3 years ago

What is the momentum of an object that has a mass of 0.03 grams and a velocity of 1200 m/s2

ruslelena [56]

answer is 36

because the formulae of momentum is

mass×velocity

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3 years ago

After pollination where does a seed grow in a flower?

ozzi

Answer:

the center of the flower or the ovary

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4 years ago

A transformer has 1000 turns in the primary coil and 100 turns in the secondary coil. If the primary coil is connected to a 120

Marianna [84]

Answer:

The voltage on the secondary is 12 V while the current is 0.5 A.

Explanation:

A transformer works by changing the level of the voltage and current on a circuit using a magnetic field and two coils. The ratio by wich they are changed is dependant on the ratio of turns between the primary and secondary of the transformer. In this case we have a ratio for the voltage of:

ratio = (turns on the secondary)/(turns on the primary)

ratio = 100/1000 = 0.1

So in this case the voltage delivered to the primary will be multiplied by 0.1. We can now calculate the voltage on the secondary:

Voltage secondary = Voltage primary* ratio = 120*0.1 = 12 V

The transformer maintains roughly the same power output on both sides, since the power output on a electric circuit is given by the product of the voltage by the current on that circuit, to maintain the same power when the voltage has been droped the current must be raised by the same ratio. So we have:

Current secondary = Current primary*(1/ratio) =0.05*(1/0.1) = 0.5 A

6 0

3 years ago

g Determine the magnetic field midway between two long straight wires 2.0 cm apart in terms of the current I in one when the oth

MrMuchimi

Answer:

$B_{net}$ = $(2\times 10^{-5}T/A).(I-25A)$

Explanation:

Given :

the distance between the two wires is = 2 cm

= 0.02 m

Current in one of the wire, I = 25 A

Now we know that, when the direction of the current is same that is in one direction , the magnetic field at the mid point in between the two wires will oppose each other.

Therefore we know that

$B_{net}$ = $\frac{\mu _{o}\times I_{1}}{2\pi r_{1}}-\frac{\mu _{o}\times I_{2}}{2\pi r_{2}}$

= $\frac{(4\pi \times 10^{-7}T.m/A)}{2\pi (1\times 10^{-2}m)}\times (I-25A)$

= $(2\times 10^{-5}T/A).(I-25A)$

4 0

3 years ago