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Arte-miy333 [17]
2 years ago
7

A gas has a volume of 50.0 cm3 at a temperature of -73°c. what volume would the gas occupy at a temperature of -123°c if the pre

ssure stays constant?
Chemistry
1 answer:
vaieri [72.5K]2 years ago
7 0

Boyle’s Law illustrates the inverse relationship of volume and pressure. It follows the formula  p1V1 = P2V2 , where P1V1 denotes initial pressure and volume  and P2V2 denotes values of pressure and  volume.

Now, let us work out for what is asked above.
a. if the pressure is doubled
50.0 p = V x 2p 
V = 50.0 p / 2p

= 50.0 /2

= 25.0 m^3 
b. if the pressure is cut in half
50.0 p = V x p/2 
100 p = V x p 
V = 100 m^3 
c. if the pressure is tripled
50.0 p = V x 3p 
V = 50.0 p / 3p

= 50.0 /3

=16.7 m^3 

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Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

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The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

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Answer:

Explanation:

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The empirical formula is therefore,
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Dividing the numerical coefficient by the lesser number.
<em>     XeF₆</em>
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