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andrew-mc [135]
2 years ago
13

How many grams of aluminum will be deposited by 0.1F? (Al=27) a.0.3g b. 0.9. c. 9.0g. d. 2.7g​

Chemistry
1 answer:
andrezito [222]2 years ago
5 0

Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.

<h3>What is electrolysis?</h3>

Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.

The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.

Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium

0.1 F will discharge = 0.1/3 × 27 g of aluminium

mass of aluminium deposited = 0.9 g of aluminium.

Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.

Learn more about electrolysis at: brainly.com/question/26050361

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Which water source becomes polluted as it travels over the earth?
never [62]

Run off

Explanation:

Run-off are the water sources that becomes polluted as they travel over the earth surface.

Surface run-off are water that moves on the earth surface. This water is a component of the water cycle which usually originate from precipitation.

  • As the water moves through different terrain it comes into contact with different pollutants.
  • Some of these run-off travels through mines, waste disposal sites among others.
  • This leads to the contamination of this form of water as it moves through these earth materials.
  • They are the most polluted water

Learn more:

Ocean water brainly.com/question/6760255

#learnwithBrainly

8 0
3 years ago
Which of the following best describes a mole as used in chemistry?
bazaltina [42]

Answer: a very large number of objects

Explanation: Mole is the amount of objects and it has the usually Avogadro number of atoms, molecules, ions, etc in chemistry

8 0
2 years ago
A given mass of oxygen occupies 500 ml when the
matrenka [14]
Ok we can use boyle’s law (stating that P is proportional to V) to make the equation (P1V1) =(P2V2).
once we’ve done this, we can plug in the numbers:
(800•500) = (200•V2)
and then we get that
V2= 2000 ml

hope this helps!! :)
5 0
3 years ago
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
zubka84 [21]

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq)  = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g                                                                     22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

=  175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require    87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

6 0
3 years ago
In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?
Mashutka [201]
Molar mass :

NaBr = 103 g/mol

Pb(NO3)2 = 331.20 g/mol

<span><span /><span>Balanced chemical equation :

</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
      g NaBr  -------------------> 311 g Pb(NO3)2

331.20  g  =   2*103*311

331.20 g = 64066

mass ( NaBr ) =  64066 / 331.20

mass ( naBr)  = 193,43 g of NaBr

hope this helps!.


</span>
5 0
3 years ago
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