Answer:
The common oxide of nitrogen that has a positive ΔS°f is nitric oxide (NO)
Explanation:
Without reference to thermodynamic data, we have;
1) N₂ (g) + O₂ (g) ⇄ 2 NO (g)
1 unit of N₂ + 1 unit of O₂ (total of 2 units) gives 2 units of NO, (Increase of +0 disorder)
∴ΔS°f = +ve
2) 2NO + O₂ → 2NO₂
2 unit of NO + 1 unit of O₂ (total of 3 units) gives 2 units of NO₂, (Decrease of disorder)
∴ΔS°f = -ve
3) N₂ + 1/2 O₂ → N₂O
1 unit of N₂ + 1/2 unit of O₂ (total of 1+1/2 units) gives 2 units of NO₂, (Decrease of disorder)
∴ΔS°f = -ve
4) 4 NO₂ + O₂ → 2N₂O₅
4 unit of NO₂ + 1 unit of O₂ (total of 5 units) gives 2 units of N₂O₅, (Decrease of disorder)
∴ΔS°f = -ve
5) NO + NO₂ ⇄ N₂O₃
1 unit of NO + 1 unit of NO₂ (total of 2 units) gives 1 unit of N₂O₃, (Decrease of disorder)
∴ΔS°f = -ve
Therefore, the common oxide of nitrogen that has a positive ΔS°f without reference to thermodynamic data is nitric oxide NO.
Answer:
cell membrane
Explanation:
Its definitely cell membrane, because the cell membrane is the outer layer, try looking up a animal cell chart, you'll see a black line around the cell. That black line is a cell membrane, the cell membrane covers over all its contents.
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
