<u>Answer:</u> The equilibrium concentration of
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of
= 2.00 M
The given chemical equation follows:

<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of
is 0.332 M
Answer:
D. +5.7 kJ/mol
Explanation:
Molar free energy (ΔG) in the transportation of uncharged molecules as glucse through a cell membrane from the exterior to the interior of the cell is defined as:
ΔG = RT ln C in / C out
knowing R is 8,314472 kJ/molK; T is 298K Cin = 200mM and Cout = 20mM
ΔG = 5,7 kJ/mol
Right answer is:
D. +5.7 kJ/mol
I hope it helps!
The type of reaction which occurs is referred to as redox reaction. This kind of reaction involve both oxidation and reduction.
Al(s) is oxidized to alluminium ions, while cu2+ is reduced copper metal.Reduction occurs at the cathode while oxidation occurs at the anode.
Alkina metals all have the similar proteries
Answer:
B
Explanation:
The general equation for the reaction of a carboxylic acid with an alkanol to form an ester is shown below;
RCOOH + ROH ------> RCOOR + H2O
Hence; the reactant carboxylic acid can only be the compound (CH3)2-CH-CH2-COOH in accordance with the general reaction equation shown above.
Hence the reaction is;
(CH3)2-CH-CH2-COOH + CH3-CH2OH -------> CH3CH2 OCO-CH2-CH-(CH3)2