All categories

3 years ago

Suppose an atom gains an e-. What will be the overall electrical charge? What if an atom loses one e-?

2 answers:

7 0

Depending on what the original charge of the atom is, when an e- is added, that charge will drop by one. If an atom loses an e- the charge will be one higher.

Tatiana [17]3 years ago

7 0

Answer:

Atoms do not always contain the same number of electrons and protons, although this state is common. When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral. In contrast, when an atom loses or gains an electron (or the rarer case of losing or gaining a proton, which requires a nuclear reaction), the total charges add up to something other than zero. The atom is then said to be electrically charged, or "ionized". There is a major difference between the neutral state and the ionized state. In the neutral state, an atom has little electromagnetic attraction to other atoms. Note that the electric field of a neutral atom is weak, but is not exactly zero because the atom is not a point particle. If another atom gets close enough to the atom, they may begin to share electrons. Chemically, we say that the atoms have formed bonds.

You might be interested in

Which are found in the retina?

Tom [10]

Your answer is D rods and cones. Hope this helps!! :)

8 0

3 years ago

Read 2 more answers

In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to

Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

Explanation:

The measured pH of the solution = 5.153

$C_6H_5OH\rightarrow C_6H_5O^-+H^+$

Initially c

At eq'm c-x x x

The expression of dissociation constant is given as:

$K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}$

Concentration of phenoxide ions and hydrogen ions are equal to x.

$pH=-\log[x]$

$5.153=-\log[x]$

$x=7.03\times 10^{-6} M$

$K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}$

$K_a=9.34\times 10^{-11}$

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

4 0

4 years ago

What pressure is exerted by 0.883 mol N, in a 4.68 L steel container at 197.9 °C?

SpyIntel [72]

I really hope i helped you! :)

6 0

3 years ago

6. What is true about Calcium (Ca)?

sasho [114]

Answer:

C. naturally uncombined with other elements.

Explanation: Calcium is a chemical element with the symbol Ca and atomic number 20. As an alkaline earth metal, calcium is a reactive metal that forms a dark oxide-nitride layer when exposed to air.

8 0

3 years ago

How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work

strojnjashka [21]

2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!

8 0

4 years ago