Answer: The molar mass of each gas
Explanation:
Mole fraction is the ratio of moles of that component to the total moles of solution. Moles of solute is the ratio of given mass to the molar mass.
![\text{Mole fraction of solute}=\frac{\text{Moles of solute}}{\text{Total Moles}}](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20solute%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BTotal%20Moles%7D%7D)
Suppose if there are three gases A, B and C.
a) ![\text{Mole fraction of A}=\frac{\text{Moles of A}}{\text{ Moles of (A+B+C)}}](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20A%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20A%7D%7D%7B%5Ctext%7B%20Moles%20of%20%28A%2BB%2BC%29%7D%7D)
b) ![\text{Mole fraction of B}=\frac{\text{Moles of B}}{\text{ Moles of (A+B+C)}}](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20B%7D%7D%7B%5Ctext%7B%20Moles%20of%20%28A%2BB%2BC%29%7D%7D)
c) ![\text{Mole fraction of C}=\frac{\text{Moles of C}}{\text{ Moles of (A+B+C)}}](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20C%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20C%7D%7D%7B%5Ctext%7B%20Moles%20of%20%28A%2BB%2BC%29%7D%7D)
moles of solute =![\frac{\text {given mass}}{\text {Molar mass}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bgiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D)
Thus if mass of each gas is known , we must know the molar mass of each gas to know the moles of each gas.
Answer:
C meteoroid
Explanation:
Meteoroids are the only ones of the four that are under a kilometer in size.
Answer:
There is 26.58 grams of gold formed
Explanation:
Step 1: Data given
17.6 A of current are passed through a gold solution for 37.0 min
Molar mass of Au = 196.967 g/mol
Step 2: The equation
Au^3+ + 3e- → Au
Step 3: Calculate coulombs
17.6 Coulomb/s * 37.0 min * 60 sec/min = 39072 Coulombs
1 Faraday = 96500 Coulombs
Step 4: Calculate faraday
39072 Coulombs / 96500 Coulombs / Faraday = 0.40489 Faraday
Step 5: Calculate mass of gold formed
For every 3 Faraday of electricity used up , 1 mole Au is formed
0.40489 Faraday * 1 mole Au/ 3 Faraday = 0.13496 mole Au
196.967 g/mol * 0.13496 mol = 26.58 g Au
There is 26.58 grams of gold formed