NaOH has the higher melting point. The reason is since NaOH is an ionic compound, thus meaning that the intermolecular forces (the forces that hold the compound together) between Sodium+ and Hydroxide- are ionic - ionic forces. The charges keep them together. Ionic forces are ALOT stronger than other intermolecular forces such as dispersion, dipole-dipole, or even hydrogen bonding.
CH3OH (Methanol) has a lower melting point that Sodium Hydroxide since the intermolecular forces it entails are: Dispersion, dipole-dipole, and hydrogen bonding between Hydrogen and Oxygen. It will take LESS energy to break these attractions, than the energy required to break the attraction forces between the ionic compound NaOH.
NaOH + CH3OH --> CH3ONa + H2O Evaporate the solution to dryness, add more CH3OH and evaporate to dryness. you can repeat a few times to ensure the remaining solid is sodium methoxide
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
If some solution splashes out during the titration of NaOH, the volume at the end point will be wrong.
They both react with phenol. KOH is actually a slightly stronger base than NaOH. At low concentrations, like less than 0.1M, both NaOH and KOH are essentially 100% dissociated. However, at higher concentrations, a little more of the KOH is dissociated. The point is, both bases will remove virtually all of the H+ from phenol. You received some incorrect information.
.17
NaOH + CH3OH --> CH3ONa + H2O Evaporate the solution to dryness, add more CH3OH and evaporate to dryness. you can repeat a few times to ensure the remaining solid is sodium methoxide
The formula of formaldehyde is CH2O. The products formed are sodium formate (HCOONa) and methanol (CH3OH). The stoichiometric equation is then X CH2O + Y NaOH --> A HCOONa + B CH3OH. Balancing the equation makes the coefficients 2 CH2O + NaOH --> HCOONa + CH3OH.
The methoxide ion would steal a hydrogen ion from water, forming sodium hydroxide and methanol. CH3ONa + H2O --> CH3OH + NaOH
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
If some solution splashes out during the titration of NaOH, the volume at the end point will be wrong.
They both react with phenol. KOH is actually a slightly stronger base than NaOH. At low concentrations, like less than 0.1M, both NaOH and KOH are essentially 100% dissociated. However, at higher concentrations, a little more of the KOH is dissociated. The point is, both bases will remove virtually all of the H+ from phenol. You received some incorrect information.
.17
.26
-10.6 c
3.42 moles NaOH (39.998 grams/1 mole NaOH) = 137 grams NaOH
0.26
0.01 molar