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3 years ago

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When factor is change when a pure element becomes an ion?

1 answer:

nikklg [1K]3 years ago

6 0

<span>When a pure element becomes an ion, the factor that changes is the charge of the element. Either the element loses an electron to become a positively charged ion, or it loses a proton, to become a negatively charged ion.</span>

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What is an example of a soultion

Marrrta [24]

Answer:

Did you mean solution??

In chemistry, a solution is considered as a special type of homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.

Eg: sweet tea or coffee (sugar added to solution)

bleach (sodium hypochlorite dissolved in water)

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3 years ago

A 100.0 lb skier moves at 40.00miles/hour. Calc her kinetic energy.

Pepsi [2]

answer: its 7290 joules.

explanations: the first procedure is to convert 1 pound to kilogram. 1 kg = 2.205 hence given 100 lb so we cross multiply. 1 kg * 100 = 2.205 * x

hence x= 45 kg. let's convert 1 mile per hour = 0.45 metre per second we cross multiply by 40 mile per hour. x= 40 * 0.45= 18 m/s.

KE= 1/2 * 45 * (18)^2

= 1/2 * 45 * 14580

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3 years ago

Specify the number of protons, neutrons, and electrons in the neutral atom iron-56. Enter your answers as integers separated by

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3 years ago

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2 years ago

It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

Zigmanuir [339]

Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of $\frac {495.0}{6.023\times 10^{23}}\ kJ$ of energy.

Energy required = $82.18\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}$

$\lambda=\frac{19.878}{10^6\times \:82.18}$

$\lambda=2.4188\times 10^{-7}\ m$

Also,

1 m = 10⁻⁹ nm

So,

$\lambda=241.9\ nm$

6 0

3 years ago