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zlopas [31]
2 years ago
7

A 1.0 L volume of gas at 27.0°C exerts a pressure of 85.5 K PA what will the pressure be at 127°C assume constant volume

Chemistry
1 answer:
Marianna [84]2 years ago
4 0

Answer:

114 kPa

Explanation:

Using Gay-Lussac's law you get the equation \frac{P1}{T1} x \frac{P2}{T2} and converting celcius you get the final equation of \frac{85.5}{27+273}  x \frac{P2}{127+273} . After dividing 85.5 by 27+273(300) you get 0.285 and then you multiply 0.285 by 127+273 (400). You finally get 114 kPa

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Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2
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Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

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