I think it's called a Halide or a halogen
Answer:
51.34 %
Explanation:
You first need to calculate the theoretical yield of Bromobutane taking into account the amount of initial Butanol doing some stoichiometric calculations:
Then you calculate the yield giving the real amount you produced:
Answer:
12.3 L
Explanation:
Now we have that the rate of diffusion of the saturated hydrocarbon is R1
Rate of diffusion of sulphur dioxide is R2
Molar mass of hydrocarbon is M1
Molar mass of sulphur dioxide is 64 gmol-1
From Graham's law;
R1/R2 = √64/M1
2/1 =√64/M1
(2/1)^2 = (√64/M1)^2
4/1 = 64/M1
4M1 =64
M1 = 16
To obtain the number of moles of the gas;
(n*12) + (2n + 2) 1 = 16
12n + 2n + 2 = 16
14n + 2 = 16
14n = 16 - 2
n = 14/14
n = 1
Hence the hydrocarbon is CH4
Volume occupied by CH4 at STP = 22.4 L
Hence;
P1 = 1 atm
T1 = 273 K
V1 = 22.4 L
T2 = 300 K
P2 = 2 atm
V2 = ?
P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 1 * 22.4 * 300/2 * 273
V2 = 12.3 L
The solution with a pH of 3 in 100 times more acetic than the solution with a pH of 5. This is because the lower the pH goes, 10x the pH decreases for each pH "level." For example, a solution with a pH of 4 is 10x more acetic than pH of 5.