Chloride ions Cl –(aq) (from the dissolved sodium chloride) are discharged at the positive electrode as chlorine gas, Cl 2(g) sodium ions Na +(aq) (from the dissolved sodium chloride) and hydroxide ions OH –(aq) (from the water) stay behind - they form sodium hydroxide solution, NaOH(aq)
Answer:
I think the layers of the atmosphere does temperature increase with increasing height. It is affected by convection because it heats the lower atmosphere. It is affected by conduction because the is the transfer of thermal energy. I guess
Hope this help!
Bonds between two atoms that are equally electronegative are nonpolar covalent bonds
Answer : The partial pressure of
and
are, 84 torr and 778 torr respectively.
Explanation : Given,
Mass of
= 15.0 g
Mass of
= 22.6 g
Molar mass of
= 197.4 g/mole
Molar mass of
= 32 g/mole
First we have to calculate the moles of
and
.
![\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DC_2HBrClF_3%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DC_2HBrClF_3%7D%3D%5Cfrac%7B15.0g%7D%7B197.4g%2Fmole%7D%3D0.0759mole)
and,
![\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DO_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DO_2%7D%3D%5Cfrac%7B22.6g%7D%7B32g%2Fmole%7D%3D0.706mole)
Now we have to calculate the mole fraction of
and
.
![\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DC_2HBrClF_3%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%7D%7B%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%2B%5Ctext%7BMoles%20of%20%7DO_2%7D%3D%5Cfrac%7B0.0759%7D%7B0.0759%2B0.706%7D%3D0.0971)
and,
![\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DO_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DO_2%7D%7B%5Ctext%7BMoles%20of%20%7DC_2HBrClF_3%2B%5Ctext%7BMoles%20of%20%7DO_2%7D%3D%5Cfrac%7B0.706%7D%7B0.0759%2B0.706%7D%3D0.903)
Now we have to partial pressure of
and
.
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 gas
= total pressure of gas
= mole fraction of gas
![p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T](https://tex.z-dn.net/?f=p_%7BC_2HBrClF_3%7D%3DX_%7BC_2HBrClF_3%7D%5Ctimes%20p_T)
![p_{C_2HBrClF_3}=0.0971\times 862torr=84torr](https://tex.z-dn.net/?f=p_%7BC_2HBrClF_3%7D%3D0.0971%5Ctimes%20862torr%3D84torr)
and,
![p_{O_2}=X_{O_2}\times p_T](https://tex.z-dn.net/?f=p_%7BO_2%7D%3DX_%7BO_2%7D%5Ctimes%20p_T)
![p_{O_2}=0.903\times 862torr=778torr](https://tex.z-dn.net/?f=p_%7BO_2%7D%3D0.903%5Ctimes%20862torr%3D778torr)
Therefore, the partial pressure of
and
are, 84 torr and 778 torr respectively.