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lubasha [3.4K]
3 years ago
15

What kinds of charged particles can move between objects that are close together?

Chemistry
1 answer:
hammer [34]3 years ago
6 0
I think it's call a magnet
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If 0.55 g of a gas dissolved in 1.0 l of water at 20.0 kpa of pressure, how much will dissolve at 110.0 kPa of pressure?
GaryK [48]

Answer:sup

Explanation:

4 0
4 years ago
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydr
Neko [114]

Answer:

F⁻(aq) + H⁺(aq) ⇄ HF(aq)

Explanation:

When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. The corresponding molecular equation is:

KF(aq) + HCl(aq) ⇄ KCl(aq) + HF(aq)

The full ionic equation includes all the ions and the molecular species. HF is a weak acid so it exists mainly in the molecular form.

K⁺(aq) + F⁻(aq) + H⁺(aq) + Cl⁻(aq) ⇄ K⁺(aq) + Cl⁻(aq) + HF(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.

F⁻(aq) + H⁺(aq) ⇄ HF(aq)

5 0
4 years ago
What balances charges that may build up as reduction and oxidation occur in a voltaic cell?
anygoal [31]
I think the correct answer from the choices listed above is option A. It is the salt bridge that balances charges that may build up as reduction and oxidation occur in a voltaic cell. The salt bridge is to <span>maintain charge balance because the electrons are moving from one half cell to the other.</span>
6 0
3 years ago
Read 2 more answers
PLX HELP ASAP
oee [108]

Answer:

B) 10.1g

Explanation:

3 0
3 years ago
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Calculate the volume of 1.0 x 10-4 M CV+ solution that needs to be added to a 25.0 mL volumetric flask and diluted with deionize
xenn [34]

Answer:

The answers are: 2.5 ml (first part) and 3.0 ml (second part)

Explanation:

In order to calculate volumes required to prepare diluted solutions we use the following equation:

Vc x Cc = Vd x Cd

Where Vc and Cc are the volume and concentration respectively of concentrated solution (higher concentration) whereas Vd and Cd are volume and concentration of diluted solution (lower concentration).

In both problems we want to prepare a diluted solution and we know the final concentration (Cd) and final volume (Vd) and the initial concentration (Cc).

In first part, we have: Cc= 1.0 10⁻⁴ M; Vd= 25 ml; Cd= 1.0 10⁻⁵ M

Vc= Vd x Cd / Cc= (25 ml x 1.0 10⁻⁵ M)/1.0 10⁻⁴ M = 2.5 ml

Notice that Cc/Cd= 1.0 10⁻⁴ M/Cd= 1.0 10⁻⁵ M= 10 (so, we have to dilute the solution 10 times, and for this we have to take a volume 10 times lower than the final volume).

To prepare the solution, we take 2.5 ml of 1.0 10⁻⁵ M CV+, we dispense the volume in a 25 ml volumetric flask and then we add water until complete 25 ml (aproximately 22.5 ml of water).

In the second part is the same. We have: Vd= 10 ml; Cc= 1.0 10⁻⁴M; Cd= 3.0 10⁻⁵M.

Vc= Vd x Cd / Cc= (10 ml x 3.0 10⁻⁵ M)/1.0 10⁻⁴ M = 3 ml

To prepare the solution, we take 3 ml of 3.0 10⁻⁵ M CV+, we dispense the volume in a 10 ml volumetric flask and then we add water until we complete 10 ml (aproximately 7 ml of water).

6 0
3 years ago
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