Answer:
a. The second run will be faster.
d. The second run has twice the surface area.
Explanation:
The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.
The area of the 10.0 cm³-sphere is:
The area of each 1.25 cm³-sphere is:
The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²
The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00
Since the surface area is doubled, the second run will be faster.
The question above is incomplete, the full question is given below:
What additional test would be needed to establish the exact position of hydrogen in the activity series of the following elements: magnesium, zinc, lead, copper and silver.
ANSWER
The position of hydrogen on a reactivity series can be determined by its ability to displace oxygen from the oxide of the metal concerned. If hydrogen is more reactive than a metal, it will displace oxygen from the metal oxide and reduce the metal oxide to its metal. If the metal is more reactive than hydrogen, hydrogen will not be able to reduce the metal oxide to its metal.
a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
The first blank is a compound, second is a mixture
Answer:
The answer to your question is V2 = 23.52 l
Explanation:
Data
Volume 1 = V1 = 22.5 l
Pressure 1 = P1 = 734 mmHg
Volume 2 = V2 = ?
Pressure 2 = 702 mmHg
Process
To solve this problem use Boyle's law.
P1V1 = P2V2
-Solve for V2
V2 = P1V1 / P2
-Substitution
V2 = (734 x 22.5) / 702
-Simplification
V2 = 16515 / 702
-Result
V2 = 23.52 l
-Conclusion
If we diminish the pressure, the volume will be higher.
