1. 5 ethyl, 2 methyl octane
2. 1 ethyl, 2 methyl cyclopentane
3. 3,3,5,5- tetrafluoro heptane
4. 3,4-dimethyl hexene
5. 3,4-dimethyl cyclobutene
6. 3,5 diisopropyl cyclohexene
7. 3,3,4 trimethyl pentyne
8. 2,6 dibromo phenol
keep in mind that between 4-7, there could be #1 in front of the main name. for example with #4: 3,4-dimethyl-1- hexene. this honestly depends on the professor how he/she likes it. It is not necessary because if the number is not specified, it is assumed is #1
Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %
At higher temperature, and lower pressure.
4 total bonds if u have double,or triple then subtract from 4
Answer:
35.8 g
Explanation:
Step 1: Given data
Mass of water: 63.5 g
Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C
Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.
63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl
