Answer:
Because water can absorb a lot of heat without significant changes in temperature, it can prevent change in the temperature, and therefore pressure, of the compressed air.To maintain a constant temperature, we again pass the compressed air through the same chambers, this time letting it take up heat from the water
The solute is the part of the solution that dissolves in the second component (usually a fluid). Therefore, for the mentioned solution, the solute is ehyl alcohol since it is the one dissolving in water.
As for the solvent, it is the component in which the solute dissolves. In this case, it is water.
Answer:
8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt
Explanation:
Let x represent the liters of the first salt solution
Let y represent the liters of the second salt solution
x + y= 20......equation 1
x= 20-y
The first solution is 20% salt and the second solution is 45% salt
20/100x + 45/100 y= 30/100 × 20
0.2x + 0.45y= 0.3×20
0.2x + 0.45y= 6............equation 2
Substitute 20-y for x in equation 2
0.2(20-y) + 0.45y= 6
4-0.2y+ 0.45y= 6
4 + 0.25y= 6
0.25y= 6-4
0.25y= 2
y= 2/0.25
y= 8
Substitute y for 8 in equation 1
x + y= 20
x + 8=20
x= 20-8
x= 12
Hence 8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt.
It is sulfur, selenium, oxygen
Answer:
The final temperature of the given ideal diatomic gas: <u>T₂ = 753.6 K</u>
Explanation:
Given: Atmospheric pressure: P = 1.0 atm
Initial Volume: V₁ , Final Volume: V₂ = V₁ (1/10)
⇒ V₁ / V₂ = 10
Initial Temperature: T₁ = 300 K, Final temperature: T₂ = ? K
For a diatomic ideal gas: γ = 7/5
For an adiabatic process:
![\left [\frac{V_{1}}{V_{2}} \right ]^{\gamma-1 } = \frac{T_{2}}{T_{1}}](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7BV_%7B1%7D%7D%7BV_%7B2%7D%7D%20%5Cright%20%5D%5E%7B%5Cgamma-1%20%7D%20%3D%20%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D)
![\left [10 \right ]^{\frac{7}{5}-1 } = \frac{T_{2}}{300 K}](https://tex.z-dn.net/?f=%5Cleft%20%5B10%20%5Cright%20%5D%5E%7B%5Cfrac%7B7%7D%7B5%7D-1%20%7D%20%3D%20%5Cfrac%7BT_%7B2%7D%7D%7B300%20K%7D)
![\left [10 \right ]^{\frac{2}{5} } = \frac{T_{2}}{300 K}](https://tex.z-dn.net/?f=%5Cleft%20%5B10%20%5Cright%20%5D%5E%7B%5Cfrac%7B2%7D%7B5%7D%20%7D%20%3D%20%5Cfrac%7BT_%7B2%7D%7D%7B300%20K%7D)
<em><u>Therefore, the final temperature of the given ideal diatomic gas</u></em><em>:</em> T₂ = 753.6 K
