Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
<u>Explanation:</u>
In a double displacement reaction, the reactants which are involved in the reaction exchanging their ions thereby produces 2 new compounds. Here sodium chromate and lead chloride are undergoing double displacement reaction, the ions exchanges their position there by forming sodium chloride and lead chromate. So the double displacement reaction is given as,
Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
Answer:
Explanation:
Carboxylic acids is a homologous series in which the compounds contain a functional group called the carboxyl group (-COOH). The general molecular formula for carboxylic acids is CnH2n+1COOH. Carboxylic acids contain at least one carboxyl group.
Or in Polish…
Kwasy karboksylowe to homologiczna seria, w której związki zawierają grupę funkcyjną zwaną grupą karboksylową (-COOH). Ogólny wzór cząsteczkowy dla kwasów karboksylowych to CnH2n + 1COOH. Kwasy karboksylowe zawierają co najmniej jedną grupę karboksylową.
The forces between particles are called intermolecular forces. A strong intermolecular force means that the particles are tightly paced and is associated with the solid phase. Moderate intermolecular force is associated with the liquid state and little to no intermolecular force is associated with the gaseous state. Temperature has a direct effect on the state of matter in which the substance exists has. Generally speaking, a rise in tempreature changes a substance from the solid to liquid phase and from liquid to gaseus phase. The reverse is true, if the temperature lowers then the substance will go from gas to liquid and liquid to solid. It is important to not that temperature affects intermolecular forces. As the temperature increases then the individual particles become excited and gain enough energy to over the intermolecular forces and so the particles seperate from each other.
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
0.511
Now using the equilibrium constant 
![K_c = \dfrac{[F6P]}{[G6P]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5BG6P%5D%7D)
![0.511 = \dfrac{[F6P]}{[2.65]}](https://tex.z-dn.net/?f=0.511%20%3D%20%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5B2.65%5D%7D)
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
