Answer:
It basically messes up the results
Explanation:
Pen ink consists of resins, pigments and other colouring dyes dissolved in appropriate solvents like propylene glycol, propyl alcohol and some other ethers. If the ball point pen is used to mark on the chromatography paper then these pigments will also move along with the solvent and interfere with the spots of our analyte.
If you use a ball point pen when doing a chromatogram, then the ink would separate as it is a mixture and run down the paper.
Graphite, or pencil lead however, is not an organic material and therefore will not be affected by common organic solvents used for thin-layer chromatography. Pen ink on the other hand will be readily absorbed by the solvent and will move up the plate.
Answer:
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¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2
2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}}
1molBa(OH)
2
1molCO
2
= 2 moles CO₂
Explanation:
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An Exothermic reaction releases energy into the surroundings and so the products have more potential energy then the reactants. The enthalpy change is a negative value. Whereas, an endothermic reaction involves the absorption of energy into the system and so the reactants have more potential energy than the products. The enthalpy change is a positive value. This is clearly represented in energy profile diagrams.
Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Explanation :
Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.
In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
We are taking the value of standard reduction potential form the standard table.
![E^0_{[Ag^{+}/Ag]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.80V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.
The balanced two-half reactions will be,
Oxidation half reaction (Anode) : 
Reduction half reaction (Cathode) : 
Thus the overall reaction will be,

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.