Answer:
The answer to your question is 8,32 L
Explanation:
Data
Pressure 1 = 104.5 kPa
Volume 1 = 3.4 L
Pressure 2 = 42.7 kPa
Volume 2 = ?
Formula
- To solve this problem use the Boyle's law.
P1V1 = P2V2
- Solve for V2
V2 = P1V1 / P2
- Substitution
V2 = (104.5)(3.4) / 42.7
-Simplification
V2 = 355.3 / 42.7
- Result
V2 = 8.32 L
To get moles. divide mass by molar mass.Molar mass of
Na is 23
and for Cl is 35.5.
the total molar mass of NaCl is 23+35.5 = 58.5mol/gUse the mass and divide by this number30.22g divide by 58.5mol/g and you will get 0.5166 mole.
Since the molecule has 1 Na to 1 Cl, and that the number of moles for NaCL is 0.5166. All of them would be 0.5166molesNa = 0.5166 x 1 = 0.5166molesCl = 0.5166 x 1 = 0.5166moles
to get number of atoms. Multiply your mole by Avogadro number which is 6.022x10^23Na = 0.5166 x 6.022E23 = 3.111x10^23Cl = 0.5166 x 6.022E23 = 3.111x10^23
Answer:
[Cl2] equilibrium = 0.0089 M
Explanation:
<u>Given:</u>
[SbCl5] = 0 M
[SbCl3] = [Cl2] = 0.0546 M
Kc = 1.7*10^-3
<u>To determine:</u>
The equilibrium concentration of Cl2
<u>Calculation:</u>
Set-up an ICE table for the given reaction:
I 0 0.0546 0.0546
C +x -x -x
E x (0.0546-x) (0.0546-x)
![Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BSbCl3%5D%5BCl2%5D%7D%7B%5BSbCl5%5D%7D%5C%5C%5C%5C1.7%2A10%5E%7B-3%7D%20%3D%5Cfrac%7B%280.0546-x%29%5E%7B2%7D%20%7D%7Bx%7D%20%5C%5C%5C%5Cx%20%3D%200.0457%20M)
The equilibrium concentration of Cl2 is:
= 0.0546-x = 0.0546-0.0457 = 0.0089 M
