Answer:
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Explanation:
We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.
The final temperature is: 75.11 °C.
The work done at constant pressure, W=nR(T₂-T₁)
n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).
W=2.4 × 10³ Joule (Given)
From the expression,
(T₂-T₁)=
(T₂-T₁)= 
(T₂-T₁)= 48.11
T₂=300+48.11=348.11 K= 75.11 °C
Final temperature is 75.11 °C.
Answer:
506.912 L
Explanation:
From the question given above, the following data were obtained:
Number of mole of O₂ = 22.63 moles
Volume of O₂ =?
Recall:
1 mole of a gas occupy 22.4 L at STP.
With the above information, we obtained the volume occupied by 22.63 moles of O₂ as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 22.63 moles of O₂ will occupy = 22.63 × 22.4 = 506.912 L at STP.
Thus, 22.63 moles of O₂ is equivalent to 506.912 L.
Answer:
24.03 J/mol.ºC
Explanation:
For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.
-Qal = Qw + Qc (minus signal represents that the heat is lost)
-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc
Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC
-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)
1971.15Cal = 1699.10 + 58.22
1971.15Cal = 1757.32
Cal = 0.89 J/g.ºC
The molar mass of Al is 27 g/mol
Cal = 0.89 J/g.ºC * 27 g/mol
Cal = 24.03 J/mol.ºC
Answer:
The answer is
<h2>0.784 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
<h3>
</h3>
From the question
mass of liquid = 9.8 g
volume = 12.5 mL
The density is
We have the final answer as
<h3>0.784 g/mL</h3>
Hope this helps you
