For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
Use equation
number of moles= sample mass/molar mass
Answer:
The Retention factor (rf) value is = 0.2
Explanation:
- Retention factor (Rf) is factor used substances that could be separated using Chromatography. Retention factor determines how fast the component can move on the chromatogram (stationary phase) after elution. Elution occurs when mobile phase (solvent) moves across the stationary phase when the solute has been spotted on the origin.
- Retention factor (Rf) ranges from value between 0 and 1. The closer the value to 1, the faster it can move upon elution. Rf can be calculated.
- Rf value = distance moved by the solute / distance moved by the solvent
= 0.40cm / 2.00cm
= 0.2
Answer:
The value of Kc for the reaction is 3.24
Explanation:
A reversible chemical reaction, indicated by a double arrow, occurs in both directions: reagents transforming into products (
direct reaction) and products transforming back into reagents (inverse reaction)
Chemical Equilibrium is the state in which direct and indirect reactions have the same reaction rate. Then taking into account the rate constant of a direct reaction and its inverse the chemical constant Kc is defined.
Being:
aA + bB ⇔ cC + dD
where a, b, c and d are the stoichiometric coefficients, the equilibrium constant with the following equation:
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.
Then, in the reaction 3A(g) + 2B(g) ⇔ 2C(g), the constant Kc is:
![Kc=\frac{[C]^{2} }{[A]^{3} *[B]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7B2%7D%20%7D%7B%5BA%5D%5E%7B3%7D%20%2A%5BB%5D%5E%7B2%7D%20%7D)
where:
- [A]= 0.855 M
- [B]= 1.23 M
- [C]= 1.75 M
Replacing:

Solving you get:
Kc=3.24
<u><em>The value of Kc for the reaction is 3.24</em></u>