It was J. J. Thomson (Joseph John Thomson) i believe!!
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.
Answer:
I think it is air particles and properties such as sollid , liquid and gas
Explanation:
The formula for beryllium chlorite is Be(ClO2)2.
The beryllium cation has a charge of 2+ and a chlorite ion has a charge of 1-. Using the crisscross method, we move the charge of the beryllium cation to the subscript position of the chlorite ion and the charge of the chlorite ion to the subscript position of the beryllium cation. So, we get Be1(ClO2)2. The ratio one-to-two cannot be reduced further. Since the subscript of Be is 1, we drop it and get Be(ClO2)2.