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Dmitry [639]
3 years ago
15

4.A 100 L sample of gas is at a pressure of 80 kPa and a temperature of 200 K. What volume does the same

Chemistry
1 answer:
dexar [7]3 years ago
8 0

Answer:

V₂ = 107.84 L

Explanation:

Given data:

Initial volume = 100 L

Initial pressure = 80 KPa (80/101 =0.79 atm)

Initial temperature = 200 K

Final temperature =273 K

Final volume = ?

Final pressure = 1 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁T₂  /T₁P₂

V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm

V₂ =21567 atm.L.K /200 K.atm

V₂ = 107.84 L

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Missing particles: ^{0}_{1}e^{+} (a positron) and an electron neutrino \nu_{\rm e}.

The nuclear equation would be:

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The mass number of a particle is the number on the top-right corner of its symbol.

The atomic number of a particle is the number on the lower-right corner of its symbol.

The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by 1.

A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron ^{0}_{1}e^{+} and an electron neutrino \nu_e would be released. On the other hand, in a beta-minus decay, a negatively-charged electron \rm ^{0}_{1}e^{-} and an electron antineutrino \overline{\nu}_e would be released.

Electric charge needs to be conserved in nuclear reactions, including this one.

The atomic number of the \rm Ne nucleus on the left-hand side is 10, meaning that the nucleus has a charge of +10. On the other hand, the atomic number of the \rm F nucleus on the right-hand side shows that this nucleus carries a charge of only +9.

By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of +1. That rules out the possibility of the combination of one negatively-charged electron \rm ^{0}_{1}e^{-} (with a charge of -1) and an electron antineutrino \overline{\nu}_e (with no electric charge at all.)

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