Here's my best guess
the volume of the unit cell is (385*10^-12)^3=5.7066*10^-29 m^3
multiply by density to get mass
mass = (7 g/cm^3)*(100^3 cm^3 / 1^3 m^3) * 5.7066*10^-29 m^3= 3.99466*10^-22 g
covert to moles
3.99466*10^-22 g * 1 mol / 239.82 g = 1.6657 *10^-24 mol
convert to number of units
1.6657 *10^-24 mol * 6.23*10^23 units/mol = 1.04
385 pm = 3.85*10^(-8) cm
The volume of the unit cell is the cube of that, which is 5.71*10^(-23) cm^3. Since the ratio of mass to volume (i.e. the density) must be the same no matter what amount of TlCl you have, you can say:
7 = x/(5.71*10^(-23)), where x is the mass of the unit cell. Solving for x, you get 4*10^(-22) g.
The mass of a molecule of TlCl is 240 amu, which in grams is 4*10^(-22) g. The mass of the unit cell and the mass of a molecule of TlCl is the same. Therefore there is one formula unit of TlCl per unit cell.
Answer:
Explanation:
The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is aspirin. Calculate the vapor pressure of the solution at 25 °C when 14.88 grams of aspirin, C9H8O4 (180.1 g/mol), are dissolved in 269.2 grams of diethyl ether. diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol.
mol of C4H10O = mass of C4H10O / molar mass of C4H10O
= 242.1 g / 74.12 g/mol
= 3.266 mol
mol of C9H8O4 = mass of C9H8O4 / molar mass of C9H8O4
= 10.33 g / 180.1 g/mol
= 0.05736 mol
mole fraction of C4H10O,
X = mole of CHH1O0 / total mol
= (3.266)/(3.266 + 0.05736)
= 0.9827
now use:
P = Po*X
P = 463.57 * 0.9827
= 455.6 mm Hg
Answer:
39.1-32.5 and you will find your answer it always like that, you subtract your starting point from your ending point
Explanation:
Explanation:
When pH of the solution is 11.
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![11=-\log[H^+]](https://tex.z-dn.net/?f=11%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=1\times 10^{-11} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1%5Ctimes%2010%5E%7B-11%7D%20M)
..(1)
At pH = 11, the concentration of 
ions is 
.
When the pH of the solution is 6.
![pH=-\log[H^+]'](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%27)
![6=-\log[H^+]'](https://tex.z-dn.net/?f=6%3D-%5Clog%5BH%5E%2B%5D%27)
![[H^+]'=1\times 10^{-6} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%27%3D1%5Ctimes%2010%5E%7B-6%7D%20M)
..(2)
At pH = 6, the concentration of ions is 
.
On dividing (1) by (2).
![\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} M}{1\times 10^{-6}}=1\times 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BH%5E%2B%5D%27%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-11%7D%20M%7D%7B1%5Ctimes%2010%5E%7B-6%7D%7D%3D1%5Ctimes%2010%5E%7B-5%7D%20)
The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is 
.
Difference between the ions at both pH:
This means that Hydrogen ions in a solution at pH = 7 has 
ions fewer than in a solution at a pH = 6
Answer:
K = 6.5 × 10⁻⁶
Explanation:
C₅H₆O₃ ⇄ C₂H₆ + 3CO
Use PV=nRT to find the initial pressure of C₅H₆O₃
P (2.50) = (0.0493) (0.08206) (473)
P = 0.78atm
C₅H₆O₃ ⇄ C₂H₆ + 3CO
0.78atm 0 0
0.78 - x x 3x
1.63atm = 0.78 - x + x + 3x
P(total) = 0.288atm
C₅H₆O₃ = 0.78 - 0.288
= 0.489atm
C₂H₆ = 0.288atm
CO = 0.846atm
= 0.379
= 6.5 × 10⁻⁶
