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3 years ago

When an atom gains an electron to become an ion what happens to its size?

1 answer:

5 0

It stays roughly the same size. Electrons have a barely imperceptible mass so the overall mass of the atom is changed very little.

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After isolating the pure crystals on a Hirsch funnel, wash them with solvent, but they dissolve and disappear into the funnel. W

AlladinOne [14]

Answer:

The pure crystals were soluble in the wash solvent.

Explanation:

The pure crystals on the funnel dissolved because they were soluble in the solvent used for washing. They were carried into the filter flask by the solvent.

3 0

3 years ago

Enter the electron configuration for the ion most likely formed by phosphorus. Express your answer in the order of orbital filli

Bingel [31]

Answer:

[Ne] 2s2 2p3

Explanation:

Phosphorus will most likely have an ion that will be 3- because it wants to have a full outer shell. Thus, the elctron configuration is: 1s2 2s2 2p6 3s2 3p3.

4 0

3 years ago

How many grams of water at 50◦c must be added to 16 grams of ice at −12◦c to result in only liquid water at 0◦c?

melisa1 [442]

When water at 50 C is added to ice at -12 C, heat is transferred from hot water to ice.

- Heat given out by water = Heat absorbed by ice

Calculating the heat released by hot water:

$q = m Cwater$ ΔT $q = m (4.184\frac{J}{(g.^{0}C)} )$ $(0^{0}C-50^{0}C)$

Calculating heat absorbed by 16 g of ice: Ice at $-12^{0}C$ is converted to ice at $O^{0}C$ and then ice at $O^{0} C$ to water at $0^{0}C$

$q = m Cice$ ΔT + $m (Heat of fusion)$

$q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C))$ + $16 g (333.55\frac{J}{g})$

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-( $m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J$

m = 27.4 g

Therefore, 27.4 g water at $50^{0}C$ must be added to 16 g of ice at $-12^{0}C$ to convert to liquid water at $0^{0}C$

4 0

3 years ago

Use the idea of diffuseition to explain how the smell of perfume travel

pogonyaev

When perfume is sprayed in a room the particles of perfume diffuse with the particles in the air.

6 0

3 years ago

How does a buffer resist change in ph upon addition of a strong acid?

lora16 [44]

Any buffer exists in this equilibrium

HA <=> $H^+ + A^-$
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base ( $A^-$ )

When a strong acid is added, it reacts with the large reservoir of the conjugate base ( $A^-$ ) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.

8 0

3 years ago