Answer : The partial pressure of
is 102.3 mmHg.
Explanation :
As per question,
Mass of
= 62.9 g
Mass of
= 33.2 g
Molar mass of
= 18 g/mole
Molar mass of
= 62 g/mole
First we have to calculate the moles of
and
.
![\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{62.9g}{18g/mole}=3.49mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DH_2O%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DH_2O%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DH_2O%7D%3D%5Cfrac%7B62.9g%7D%7B18g%2Fmole%7D%3D3.49mole)
![\text{Moles of }HOCH_2CH_2OH=\frac{\text{Mass of }HOCH_2CH_2OH}{\text{Molar mass of }HOCH_2CH_2OH}=\frac{33.2g}{62g/mole}=0.536mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DHOCH_2CH_2OH%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DHOCH_2CH_2OH%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DHOCH_2CH_2OH%7D%3D%5Cfrac%7B33.2g%7D%7B62g%2Fmole%7D%3D0.536mole)
Now we have to calculate the mole fraction of
and
.
![\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{3.49}{3.49+0.536}=0.867](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DH_2O%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH_2O%7D%7B%5Ctext%7BMoles%20of%20%7DH_2O%2B%5Ctext%7BMoles%20of%20%7DHOCH_2CH_2OH%7D%3D%5Cfrac%7B3.49%7D%7B3.49%2B0.536%7D%3D0.867)
![\text{Mole fraction of }HOCH_2CH_2OH=\frac{\text{Moles of }HOCH_2CH_2OH}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{0.536}{3.49+0.536}=0.134](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DHOCH_2CH_2OH%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DHOCH_2CH_2OH%7D%7B%5Ctext%7BMoles%20of%20%7DH_2O%2B%5Ctext%7BMoles%20of%20%7DHOCH_2CH_2OH%7D%3D%5Cfrac%7B0.536%7D%7B3.49%2B0.536%7D%3D0.134)
Now we have to partial pressure of
.
According to the Raoult's law,
![p^o=X\times p_T](https://tex.z-dn.net/?f=p%5Eo%3DX%5Ctimes%20p_T)
where,
= partial pressure of water vapor
= total pressure of gas
= mole fraction of water vapor
![p_{H_2O}=X_{H_2O}\times p_T](https://tex.z-dn.net/?f=p_%7BH_2O%7D%3DX_%7BH_2O%7D%5Ctimes%20p_T)
![p_{H_2O}=0.867\times 118.0mmHg=102.3mmHg](https://tex.z-dn.net/?f=p_%7BH_2O%7D%3D0.867%5Ctimes%20118.0mmHg%3D102.3mmHg)
Therefore, the partial pressure of
is 102.3 mmHg.
The expansion diameters of the steel cylinder is 0.3mm
The expansion stress on the steel cylinder is 150MPa
<h3>Diameter of thin walled cylinder </h3>
t = Dp / 2s
low pressure cylinder of cast iron used for certain engine
t = Dp / 2500 + 0.3
wall thickness = 0.3 mm
p = internal pressure
d = inside diameter of cylinder
t = wall thikness
s = allowable tensile stress
<h3>The stress of the cylinder is</h3>
The pressure in a thin walled tube with diameter 0.3 m and thickness 0.001 m is 1000 kPa (10 bar).
The hoop stress can be calculated
σh = (1000 kPa) (0.3 m) / (2 (0.001 m))
= 150000 kPa
= 150 MPa
Hence , The diameter of the thin walled cylinder is 0.3mm
The stress of the thin walled cylinder is 150 MPa
Learn more about the thin walled cylinder on
brainly.com/question/14686675
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9.00g/1hr * 1kg/100g * 1hr/60min = 0.00015kg/min or 1.5 * 10^-4kg/min.
Answer: Laws have been passed banning the production of a living copy of a person.
i just took the test so i know the answer is correct.
Answer:
Because the atmospheric pressure at higher altitudes is lower than at sea level.
Water, or any liquid for that matter, boils when its vapor pressure equals the external pressure above the liquid.