Answer:
B. Poor conductor.
Explanation:
It cannot be A, as only 1 metal is not solid at room temp.
It cannot be C, as most metals are ductile.
It cannot be D, as most metals are malleable.
This leaves B, which is not true about metals, as a lot are very good conductors.
Answer:
The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
Explanation:
Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen using Lead (IV) oxide as a catalyst.
- The catalyst surface area is directly proportional to the reaction rate
- So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.
2. Also, Removing lead (IV) oxide from the reaction mixture the reaction rate decreased because as the catalyst is removed.
3. Using 50 cm³ of hydrogen peroxide doesn't affect the rate because the concentration of the reactant doesn't change.
4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased
So, The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
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.
Au2S is =<u><em> 426 g/mol</em></u> for molar mass
