For the answer to the question above, <span>The formula for freezing point depression is </span>
<span>ΔTf = mkfi </span>
<span>kf is the freezing point constant </span>
<span>i is the Van't Hoff factor which in this case is 1 </span>
<span>m is molality (moles of solute/kg of solvent) </span>
<span>ΔTf is temperature change </span>
<span>ΔTf is 2.17 °C, the molality is the amount of solute Quinine </span>
<span>in the solvent cyclohexane. We cannot calculate moles therefore we need to substitute moles with g/mm. </span>
<span>moles = g/mm so molality=(g/mm)/kg </span>
<span>molality = (0.845/mm)/0.025 = 33.8/mm </span>
<span>2.17 = 33.8/mm(20.8) rearrange </span>
<span>mm = (33.8/2.17)(20.8) = 324g/mol</span>
Percent yield is calculated by obtaining the ratio of the actual yield and the theoretical yield times 100.
Actual yield = 1500 kg
Theoretical yield = 1500 ( 1 / 30.00 ) ( 2 / 2 ) ( 46.006 / 1 ) = 2300.3 kg
Percent yield = 1500 / <span>2300.3</span> x 100 =<span> 65.21 %
Therefore, the correct answer is the last option. Hope this answers the question.</span>
There are four quantum numbers: the n, l, m and ms. When the orbital is 2p, it means that n=2 and l = 1. The n number takes the whole number before the sub-orbitals. Then, the designation for the p orbital is l=1. Now, the possible values if m is a range from -l to l. H<em>ence, in this case, that would be: -1, 0, 1.</em>
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
![V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}](https://tex.z-dn.net/?f=V_%7BL%7D%3D%5Cfrac%7B4%7D%7B3%7D%2Api%2A%7Br_%7BL%7D%7D%5E%7B3%7D)
![10=\frac{4}{3}*pi*{r_{L}}^{3}](https://tex.z-dn.net/?f=10%3D%5Cfrac%7B4%7D%7B3%7D%2Api%2A%7Br_%7BL%7D%7D%5E%7B3%7D)
![{r_{L}}^{3}=\frac{7.5}{pi}](https://tex.z-dn.net/?f=%7Br_%7BL%7D%7D%5E%7B3%7D%3D%5Cfrac%7B7.5%7D%7Bpi%7D)
![r_{L}=1.337cm](https://tex.z-dn.net/?f=r_%7BL%7D%3D1.337cm)
Step 2: Determine surface area of large sphere
![A_{L}=4*pi*{r_{L}}^{2}](https://tex.z-dn.net/?f=A_%7BL%7D%3D4%2Api%2A%7Br_%7BL%7D%7D%5E%7B2%7D)
![A_{L}=4*pi*1.337^{2}](https://tex.z-dn.net/?f=A_%7BL%7D%3D4%2Api%2A1.337%5E%7B2%7D)
![A_{L}=22.463 cm^{2}](https://tex.z-dn.net/?f=A_%7BL%7D%3D22.463%20cm%5E%7B2%7D)
Step 3: Determine radius of small sphere
![V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}](https://tex.z-dn.net/?f=V_%7Bs%7D%3D%5Cfrac%7B4%7D%7B3%7D%2Api%2A%7Br_%7Bs%7D%7D%5E%7B3%7D)
![1.25=\frac{4}{3}*pi*{r_{s}}^{3}](https://tex.z-dn.net/?f=1.25%3D%5Cfrac%7B4%7D%7B3%7D%2Api%2A%7Br_%7Bs%7D%7D%5E%7B3%7D)
![{r_{s}}^{3}=\frac{0.9375}{pi}](https://tex.z-dn.net/?f=%7Br_%7Bs%7D%7D%5E%7B3%7D%3D%5Cfrac%7B0.9375%7D%7Bpi%7D)
![r_{s}=0.668cm](https://tex.z-dn.net/?f=r_%7Bs%7D%3D0.668cm)
Step 4: Determine surface area of small sphere
![A_{s}=4*pi*{r_{s}}^{2}](https://tex.z-dn.net/?f=A_%7Bs%7D%3D4%2Api%2A%7Br_%7Bs%7D%7D%5E%7B2%7D)
![A_{s}=4*pi*0.668^{2}](https://tex.z-dn.net/?f=A_%7Bs%7D%3D4%2Api%2A0.668%5E%7B2%7D)
![A_{s}=5.607 cm^{2}](https://tex.z-dn.net/?f=A_%7Bs%7D%3D5.607%20cm%5E%7B2%7D)
Step 5: Determine total surface area of 8 small spheres
![A_{S}=8*A_{s}](https://tex.z-dn.net/?f=A_%7BS%7D%3D8%2AA_%7Bs%7D)
![A_{S}=8*5.607](https://tex.z-dn.net/?f=A_%7BS%7D%3D8%2A5.607)
![A_{S}=44.856 cm^{2}](https://tex.z-dn.net/?f=A_%7BS%7D%3D44.856%20cm%5E%7B2%7D)
- 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:
2160 times 4 would be the answer
Explanation:
Took it