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
Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles
Answer:
Vapour pressure of benzene over the solution is 253 torr
Explanation:
According to Raoult's law for a mixture of two liquid component A and B-
vapour pressure of a component (A) in solution = 
vapour pressure of a component (B) in solution = 
Where 
are mole fraction of component A and B in solution respectively
are vapour pressure of pure A and pure B respectively
Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr
So, vapour pressure of benzene in solution = 
= 253 torr
When a bond is broken that should be a type of reaction. When bonds are broken sometimes heat is released.
