Answer:
![T_b=239.7K=-33.49\°C](https://tex.z-dn.net/?f=T_b%3D239.7K%3D-33.49%5C%C2%B0C)
Explanation:
Hello!
In this case, since the relationship between entropy and enthalpy for any process is defined below:
![S=\frac{H}{T}](https://tex.z-dn.net/?f=S%3D%5Cfrac%7BH%7D%7BT%7D)
For the vaporization of ammonia or any liquid, we can write:
![\Delta S_{vap}=\frac{\Delta H_{vap}}{T_{vap}}](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bvap%7D%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BT_%7Bvap%7D%7D)
In such a way, solving the temperature of vaporization, or boiling point, we have:
![T_{vap}=\frac{\Delta H_{vap}}{\Delta S_{vap}}](https://tex.z-dn.net/?f=T_%7Bvap%7D%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B%5CDelta%20S_%7Bvap%7D%7D)
Plugging in the given enthalpy and entropy of vaporization we obtain:
![T_{vap}=T_b=\frac{23350\frac{J}{mol} }{97.43\frac{J}{mol*K}} \\\\T_b=239.7K=-33.49\°C](https://tex.z-dn.net/?f=T_%7Bvap%7D%3DT_b%3D%5Cfrac%7B23350%5Cfrac%7BJ%7D%7Bmol%7D%20%7D%7B97.43%5Cfrac%7BJ%7D%7Bmol%2AK%7D%7D%20%5C%5C%5C%5CT_b%3D239.7K%3D-33.49%5C%C2%B0C)
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Answer:
5 or 5.0
Explanation:
because in order to find the volume yo divide mass and density
so 50.0 or 50 divided by 10.0 or 10 is 5
hope this helped :)
Answer:
Percent yield = 90.9%
Explanation:
Given data:
Mass of CaCO₃ = 50.0 g
Mass of CO₂ produced = 20.0 g
Percent yield = ?
Solution:
Chemical equation:
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 100.1 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of CO₂ with CaCO₃.
CaCO₃ : CO₂
1 : 1
0.5 : 0.5
Mass of CO₂: Theoretical yield
Mass = number of moles × molar mass
Mass = 0.5 mol × 44 g/mol
Mass = 22 g
Percent yield:
Percent yield = ( actual yield / theoretical yield ) × 100
Percent yield = (20.0 g/ 22.0 g) × 100
Percent yield = 0.909 × 100
Percent yield = 90.9%
Glucose freezes at -3.75 °C and freezing point of pure water is 0 °C. Thus, depression in freezing point or
can be calculated as follows:
![\Delta T_{f}=0-(-3.75)^{o}C=3.75^{o}C](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%7D%3D0-%28-3.75%29%5E%7Bo%7DC%3D3.75%5E%7Bo%7DC)
It is related to molal concentration as follows:
![\Delta T_{f}=k_{f}m](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bf%7D%3Dk_%7Bf%7Dm)
Here,
is freezing point depression constant.
Rearranging to calculate molal concentration:
![m=\frac{\Delta T_{f}}{k_{f}}=\frac{3.75 ^{o}C}{1.86^{o}C/m}=2.02 m](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%5CDelta%20T_%7Bf%7D%7D%7Bk_%7Bf%7D%7D%3D%5Cfrac%7B3.75%20%5E%7Bo%7DC%7D%7B1.86%5E%7Bo%7DC%2Fm%7D%3D2.02%20m)
Therefore, molal concentration of glucose in the solution will be 2.02 m.
Answer:
The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0.