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Alex17521 [72]
2 years ago
15

27 Points Please Help ASAP!! View Attached Image!! Will Mark Brainliest If All Are Answered!!

Chemistry
1 answer:
zvonat [6]2 years ago
3 0

<em>Hello Human!!</em>

<em>The answer is below!!</em>

<em>-------------------------------------------------------------------------------------------------------</em>

<em>14. </em><em>The answer is </em><em>C. </em><em>Mass</em>

<em>15. </em><em>I think the right answer is </em><em>C. </em><em>Melting point</em>

<em>16. </em><em>I don't know this is right.... </em><em>D. </em><em>Ability to oxidize </em>

<em>17. </em><em>I also don't know this right as well... So </em><em>A. </em><em>Generally, a gas is in the gaseous state at room temperature and a vapor is in either the  solid state or liquid state at room temperature.</em>

<em>P.S </em><em>Tell me if this wrong....</em>

<em />GoodLuck!!<em />

<em>#</em>Be<em> </em>Bold<em />

<em>#</em>LearnWithBrainlyAlways<em />

<em />_{Itbrazts}<em />

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Using the law of conservation of energy what is the kinetic energy at D
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2 years ago
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}}

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

6 0
2 years ago
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