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3 years ago

In the Energy and Specific Heat lab, how should the water bath be stabilized over the heat source? Select one:

Chemistry

2 answers:

Reika [66]3 years ago

5 0

Answer:

Explanation:

butalik [34]3 years ago

3 0

Answer:

Explanation:

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A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

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Answer:

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4 years ago

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