How long will it take to deposit 6.32 g of copper from a CuSO4(aq) solution using a current of 0.554 amps

34672.96 s

Explanation:

m = Mass of copper = 6.32 g

M = Molar mass of copper = 63.5 g/mol

F = Faraday constant = 96500 C/mol

The electrode equation would be

Number of electrons = 2 = e

I = Current = 0.554 A

t = Time taken

Charge would be

Charge is given by

Time taken to deposit the copper is 34672.96 s.

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.