All they do is simply expand when you inhale and contract when you exhale.
<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol
<u>Explanation:</u>
Rate of a gas is defined as the amount of gas displaced in a given amount of time.
![\text{Rate}=\frac{V}{t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3D%5Cfrac%7BV%7D%7Bt%7D)
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:
![\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20effusion%7D%5Cpropto%20%5Cfrac%7B1%7D%7B%5Csqrt%7B%5Ctext%7BMolar%20mass%20of%20the%20gas%7D%7D%7D)
So,
![\left(\frac{\frac{V_{X}}{t_{X}}}{\frac{V_{O_2}}{t_{O_2}}}\right)=\sqrt{\frac{M_{O_2}}{M_{X}}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7B%5Cfrac%7BV_%7BX%7D%7D%7Bt_%7BX%7D%7D%7D%7B%5Cfrac%7BV_%7BO_2%7D%7D%7Bt_%7BO_2%7D%7D%7D%5Cright%29%3D%5Csqrt%7B%5Cfrac%7BM_%7BO_2%7D%7D%7BM_%7BX%7D%7D%7D)
We are given:
Volume of unknown gas (X) = 1.0 L
Volume of oxygen gas = 1.0 L
Time taken by unknown gas (X) = 105 seconds
Time taken by oxygen gas = 31 seconds
Molar mass of oxygen gas = 32 g/mol
Molar mass of unknown gas (X) = ? g/mol
Putting values in above equation, we get:
![\left(\frac{\frac{1.0}{105}}{\frac{1.0}{31}}\right)=\sqrt{\frac{32}{M_X}}\\\\M_X=367.12g/mol](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7B%5Cfrac%7B1.0%7D%7B105%7D%7D%7B%5Cfrac%7B1.0%7D%7B31%7D%7D%5Cright%29%3D%5Csqrt%7B%5Cfrac%7B32%7D%7BM_X%7D%7D%5C%5C%5C%5CM_X%3D367.12g%2Fmol)
Hence, the molar mass of unknown gas is 367.12 g/mol
Answer:
A liquid, at any temperature, is in equilibrium with its own steam. This means that on the surface of the liquid or solid substance, there are gaseous molecules of this substance. These molecules exert a pressure on the liquid phase, a pressure known as vapor pressure.
In chemistry, when we talk about dry basis, we talk about a state in which the presence of water in a gaseous state is denied for the calculation. So vapor pressure equals zero.
When we talk about the wet basis, the presence of water in the steam is considered for the calculation, which normally is expressed as a percentage or moisture.
In summary, for a gas mixture steam:
- For dry basis, we just have <em>component A, component B....</em>
- For wet basis, we have <em>water vapor, component A, component B...</em>
So, in wet basis we have an extra component (water).
Assuming we only have 2 components in our steam, and being X the molar fraction of eact component:
- For dry basis: Xa + Xb = 1................................. Xa = 1 - Xb
- For wet basis: Xa + Xb + Xwater = 1 .............Xa = 1 - Xwater - Xb
For dry basis the mole fraction of A it is obtained by subtracting the molar fraction of B from one. And for wet basis, we have to substract the molar fraction of B <u>AND </u>the molar fraction of water vapor. So, logically, the mole fraction Xa will be less for wet basis.
D = m / V
D = 2790 g / 205 mL
D = 13.60 g/mL
Two naturally occurring isotopes of antimony are shown in the pic with their abundance