Answer:
The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
Explanation:
A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.
Step 1: Determine radius of large sphere




Step 2: Determine surface area of large sphere



Step 3: Determine radius of small sphere




Step 4: Determine surface area of small sphere



Step 5: Determine total surface area of 8 small spheres



- Surface area of 1 large sphere
- Surface area of 8 small spheres
Options:
- The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
- The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
- The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
- The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
- The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
- The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm
Answer:
0.14 M
Explanation:
To determinate the concentration of a new solution, we can use the equation below:
C1xV1 = C2xV2
Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL
1.50x55.0 = C2x278
C2 = 0.30 M
The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL
Then,
0.30x139 = C2x294
C2 = 0.14 M