Gasoline, kerosene, and lighter fluid.
In oil and gas industry:
When crude oil get extracted from well, salt water and some other stuff needs to be removed before oil can be sued in the car
Answer:
The correct answer is 169.56 g/mol.
Explanation:
Based on the given information, the mass of Ag deposited is 1.24 g, and the mass of unknown metal X deposited in another cell is 0.650 g. The number of moles of electrons can be determined as,
= 1.24 g Ag * 1mol Ag/107.87 g/mol Ag * 1 mol electron/1 mol Ag ( the molecular mass of Ag is 107.87 g/mol)
= 0.0115 mole of electron
The half cell reaction for the metal X is,
X^3+ (aq) + 3e- = X (s)
From the reaction, it came out that 3 faraday will reduce one mole of X^3+.
The molar mass of X will be,
= 0.650 g/0.0115 *3 mol electron/1 mol
= 56.52 * 3
= 169.56 g/mol
Answer: Hmmmmm that's crazy....
There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.
Fraction remaining (FR) = 0.5n
n = number of half lives that have elapsed
In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).
0.0156 = 0.5n
log 0.0156 = n log 0.5
-1.81 = -0.301 n
n = 6.0 half lives have elapsed
Explanation:
Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)
