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3 years ago

Determine the number of moles of Cr atoms in 50.00grams of Cr2Cl3

1 answer:

6 0

We know from the formula of Cr₂Cl₃, that each mole of the compound consists of 2 moles of Cr. Thus,
moles of Cr = moles of Cr₂Cl₃ x 2
= 0.3155 x 2

= 0.631 mole of Cr

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Use the IUPAC nomenclature rules to give the name for this compound - Al2S3.

jonny [76]

Aluminium Sulfide
According to rules the positive specie is named first and the negative specie is named last.

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3 years ago

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Which is an example of poor safety practices when working outdoors

kati45 [8]

Answer:

Explanation:

u can't touch a chemical with bare skin

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3 years ago

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Aluminum is manufactured using electrolysis. Carbon electrodes are used. Describe the nature of the electrolyte.

Aleksandr [31]

Answer:

The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.

Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).

However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.

3 0

3 years ago

Ammonia, NH 3 , may react with oxygen to form nitrogen gas and water. 4 NH 3 ( aq ) + 3 O 2 ( g ) ⟶ 2 N 2 ( g ) + 6 H 2 O ( l )

Alex Ar [27]

Answer:

The limiting reactant is NH₃

0.0186moles of N₂ are the one produced by the limiting reactant

0.020 moles of N₂ are the one produced by the reactant in excess

Explanation:

This is the reaction

4NH₃ + 3O₂ → 2N₂ + 6H₂O

We should calculate the moles of each reactant

Mass / Molar mass = Moles

3.55 g / 17g/m = 0.208 moles NH₃

5.33 g / 32g/m = 0.166 moles O₂

4 moles of ammonia react with 3 moles of oxygen

0.208 moles of ammonia react with (0.208 .3)/4 = 0.156 moles O₂

We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.

3 moles of O₂ react with 4 moles of NH₃

0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles

We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.

To know the moles of N₂, let's apply the Ideal Gas Law

P.V =n.R.T

1atm . 0.450L = n . 0.082 . 295K

0.450 / (0.082 .295) = 0.0186 moles

If we have 100 % yield reaction:

4 moles NH₃ make 2 moles N₂

0.208 moles NH₃ make (0.208 .2)/4 = 0.104 moles

So the % yield reaction is.

0.104 moles ___ 100%

0.0186 moles ___ 17.9%

0.0186moles of N₂ are the one produced by the limiting reactant.

3 moles of O₂ produce 2 moles N₂

0.166 moles O₂ produce (0.166 .2)/3 = 0.111 moles

Now, we apply the yield.

100% ____ 0.111 moles

17.9% = 0.020 moles

8 0

3 years ago

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Consider the following equilibrium: CO2(g) + C(graphite) 2CO(g); ΔH = 172.5 kJ The equilibrium constant for this reaction will

bezimeni [28]

Answer: Option (B) is the correct answer.

Explanation:

As the given reaction is as follows.

$CO_{2}(g) + C(graphite) \rightarrow 2CO(g)$

Equilibrium constant for this reaction will be as follows.

$K_{c} = \frac{[CO_{2}]}{[CO]^{2}}$

According to Le Chatelier's principle, when we increase the temperature then the equilibrium will shift towards the right hand side.

As a result, concentration of carbon dioxide will decrease whereas concentration of carbon monoxide will increase.

Thus, we can conclude that in the given reaction equilibrium constant for this reaction will decrease with increasing temperature.

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3 years ago