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3 years ago

Someone please help asap What general trend does electronegativity follow on the periodic table?

1 answer:

5 0

On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group. As a result, the most electronegative elements are found on the top right of the periodic table, while the least electronegative elements are found on the bottom left.

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This example provides a complete scientific description of an object in motion.

Naddika [18.5K]

"The marble moved 30 cm north in 6 seconds" is the one example among the following choices given in the question that provides a complete scientific description of an object in motion. The correct option among all the options that are given in the question is the third option or option "C". I hope the answer has helped you.

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3 years ago

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PLEASE HELP TIMED

Igoryamba

Radio waves have longer wavelengths and lower frequencies than microwaves.

infrared is longer wavelengths and lower frequencies than UV light

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3 years ago

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In the attachment there is a density column where there is colour

marusya05 [52]

Because it’s on the bottom, and to get to that part of it, think of it like that scene from moana when they go down into the realm of monsters they go through the thick parts to get to the thin parts, soo the cube wouldn’t be nearly able to reach that level

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3 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

3 years ago

two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of m

alexdok [17]

The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.

F α Qq

If the charges are now half the initial charges:

F α (1/2)Q *(1/2)q

F α (1/4)Qq

The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.

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3 years ago

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