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Ann [662]
2 years ago
5

A sample of sulfur dioxide (SO2) is initially at a temperature of 105C, a volume of 15 L, and a pressure of 1.5 atm. If the volu

me changes to 25 L and the temperature increases to 181C, what is the new pressure? Show your work.
Chemistry
1 answer:
Agata [3.3K]2 years ago
5 0

The new pressure is 1.08 atm

<h3>Gas law </h3>

From the question, we are to determine the new pressure

Using the General gas equation,

\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}

Where P₁ is the initial pressure

V₁ is the initial volume

T₁ is the initial temperature

P₂ is the final pressure

V₂ is the final volume

and T₂ is the final temperature

From the given information,

P₁ = 1.5 atm

V₁ = 15 L

T₁ = 105 C = 105 + 273.15 = 378.15 K

V₂ = 25 L

T₂ = 181 C = 181 + 273.15 = 454.15 K

P₂ = ?

Putting the parameters into the equation, we get

\frac{1.5 \times 15}{378.15} = \frac{P_{2} \times 25 }{454.15}

P_{2}= \frac{1.5 \times 15 \times 454.15}{25 \times 378.15}

P₂ = 1.08 atm

Hence, the new pressure is 1.08 atm

Learn more on Gas laws here: brainly.com/question/25736513

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The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,
Masja [62]

<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.50-0=1.50V

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

where,

E_{cell} = electrode potential of the cell = 1.23 V

E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

Hence, the concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

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Answer:

15.0L

Explanation:

p/v = constan

(9*21)/253 =(15v)/ 302

v = (9*21*302)/(15*253)

v=15.0

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Which of the following describes a strong acid?
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