Answer:
Kc = 3.1x10²
Explanation:
At equilibrium, the velocity of product formation is equal to the velocity of reactants formation. For a generic reaction, the equilibrium constant (Kc) is:
aA + bB ⇄ cC + dD
![Kc = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%2A%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%2A%5BB%5D%5Eb%7D)
Where [X] is the molar concentration of X, and the solid substances are not considered (because it's activity is 1, for the other substances, the activity is substituted for the molar concentration, which forms the equation above).
For the reaction given, let's make an equilibrium chart:
Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)
1.1*10⁻³ 8.2*10⁻⁴ 0 <em> Initial</em>
-x -x +x <em>Reacts</em> (stoichiometry is 1:1:1)
1.1*10⁻³ -x 8.2*10⁻⁴ -x x <em> Equilibrium</em>
x = 1.8*10⁻⁴ M, so the molar concentrations at equilibrium are:
[Fe⁺³] = 1.1*10⁻³ - 1.8*10⁻⁴ = 9.2*10⁻⁴ M
[SCN⁻] = 8.2*10⁻⁴ - 1.8*10⁻⁴ = 6.4*10⁻⁴ M
[FeSCN⁺²] = 1.8*10⁻⁴ M
Kc = [FeSCN⁺²]/([Fe⁺³]*[SCN⁻])
Kc = (1.8*10⁻⁴)/(9.2*10⁻⁴*6.4*10⁻⁴)
Kc = 306 = 3.1x10²
Bile salt from the gall bladder mix with fats to further break them down in a process called emulsification
Explanation
Emulsification involve breakdown of large fat globules into tiny droplet in the duodenum. This provide a large surface area on which the enzyme pancreatic lipase can act to digest the fats into fatty and qlycerol. Emulsification is assisted by action of bile salt where the secreted bile in the liver and stored in the gallbladder is released to the duodenum .
Answer:
i think it's B CI-35 only
Explanation:
hope this helps
P = nRT/V
P = 3.5 x 10^-3 x 0.082 x 298 /0.5
P = 0.171 m Hg
P = 171 mm Hg
hope this helps
<h3>Answer:</h3>
The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.
<h3>Explanation:</h3>
Lets take start with the melting point of both compounds.
n-Butane = - 140 °C
Trimethylamine = - 117 °C
Intermolecular Forces in n-Butane:
As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.
Intermolecular Forces in Trimethylamine:
Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.
