Answer:
- The answer is the concentration of an NaOH = 1.6 M
Explanation:
The most common way to solve this kind of problem is to use the formula
In your problem,
For NaOH
C₁ =?? v₁= 78.0 mL = 0.078 L
For H₂SO₄
C₁ =1.25 M v₁= 50.0 mL = 0.05 L
but you must note that for the reaction of NaOH with H₂SO₄
2 mol of NaOH raect with 1 mol H₂SO₄
So, by applying in above formula
- (C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)
- C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M
<u>So, the answer is the concentration of an NaOH = 1.6 M</u>
Answer:
Polarity results from an unequal sharing of valence electrons. In SO3 there is the sharing is equal. Therefore SO3 is a nonpolar molecule.
Explanation:
Answer:
Option-1 (Solubility and Molecular polarity) is the correct answer.
Explanation:
Thin Layer Chromatography is employed to separate a mixture of non volatile compounds. In this technique an adsorbent material like silica gel is coated on a plastic, glass or aluminium sheet. Then the mixture of compounds is applied at the bottom of sheet and the sheet is placed in the container containing a solvent system. It is observed that the solvent starts travelling upward through capillary action.
While the solvent is running the mixture of compounds starts separating from each other. This separation is due to following physical properties.
1) Solubility of Mixture in Solvent:
In a mixture those compounds which has more solubility in solvent will travel more and will give greater Rf value and the less soluble will left behind with smaller Rf value. Hence due to solubility a mixture of compounds can be separated.
2) Polarity of Molecules:
As the stationary phase (adsorbent material) is polar in nature, so in mixture those compounds which are less polar will less interact with the stationary phase and will travel more with greater Rf value, while, more polar molecules will form stronger interactions with the stationary phase, hence will travel less and therefore, will show smaller Rf values.
Answer:
300g of water because its a larger amount
Explanation:
