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: The answer is B
Explanation:
RiP BoZo. shout out to faze gabi staright up bopped potato girl.
Answer:
osmosis is a process where by solvent or fluid molecules flow from low concentration to high concentration.
So in
A.solvent will flow from B to A because
B = 3.45 M sodium bromide is more concentrated than 3.45 M calcium iodide;
b) solvent still flow from B to A because
A = 5.00 M sucrose is less concentrated than B = 3.00 M potassium nitrate
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:
ln [A] = -kt + ln [A]_o, where,
k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species
For this problem, we simply substitute the known values to the equation as in:
ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M)
We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
