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Elan Coil [88]
3 years ago
8

How many moles of gas are in a 35.0 L scuba canister if the temperature of the canister is 27.3 °C and the pressure is 200.8 ATM

Chemistry
1 answer:
Rama09 [41]3 years ago
5 0

Answer:

285.4 moles of gas are in a 35.0 L scuba canister if the temperature of the canister is 27.3 °C and the pressure is 200.8 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 200.8 atm
  • V= 35 L
  • n=?
  • R= 0.082 \frac{atm*L}{mol* K}
  • T= 27.3 C= 300.3 K (being O C= 273 K)

Replacing:

200.8 atm* 35 L= n* 0.082\frac{atm*L}{mol* K} * 300.3 K

Solving:

n=\frac{200.8 atm* 35 L}{0.082\frac{atm*L}{mol* K} * 300.3 K}

n= 285.4 moles

<u><em>285.4 moles of gas are in a 35.0 L scuba canister if the temperature of the canister is 27.3 °C and the pressure is 200.8 atm.</em></u>

<u><em></em></u>

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
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<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

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To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

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E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

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Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

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