Acids: taste sour, has a pH less than 7
bases: tastes bitter, has a pH greater than 7
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.
<h3><u>Answer;</u></h3>
Empirical formula = C₂H₃O
Molecular formula = C₁₄H₂₁O₇
<h3><u>Explanation</u>;</h3>
Empirical formula
Moles of;
Carbon = 55.8 /12 = 4.65 moles
Hydrogen = 7.04/ 1 = 7.04 moles
Oxygen = 37.16/ 16 = 2.3225 moles
We then get the mole ratio;
4.65/2.3225 = 2.0
7.04/2.3225 = 3.0
2.3225/2.3225 = 1.0
Therefore;
The empirical formula = <u>C₂H₃O</u>
Molecular formula;
(C2H3O)n = 301.35 g
(12 ×2 + 3× 1 + 16×1)n = 301.35
43n = 301.35
n = 7
Therefore;
Molecular formula = (C2H3O)7
<u> = C₁₄H₂₁O₇</u>
Explanation:
In the context, a vial which is used in store medical samples is filled with water at room temperature. And the vial is kept on a cold water. Also a water bag containing warm water is kept near the vial.
The cold water kept at the bottom of the vial is having lower kinetic energy while warm water will have higher kinetic energy than the others. Since the water in the vial is at room temperature and it is in touch with the cold blue water, the water in the vial will loose or give its temperature to the cold blue water through conduction as well as convection process since temperature always flows from a hot body towards the cold body.
On the other hand, the warm water placed next tot he vial will give its temperature to the atmosphere.