AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C°
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n = 1.45/mass weight of argon
= 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm
Answer:
A.) 1
Explanation:
Propane only exists in one conformation. It does not have enough carbons to form branches, and there are only hydrogens attached to each carbon. Furthermore, there is no way to twist the carbon or change its orientation (ex. cis- and trans-) to result in a different structure of propane. There is no other way to represent the molecule without drawing a different molecule.
Using the relative atomic weights of both copper and sulfur ie copper = 63.55 and sulfur is 32.06 so 63.55+32.06=95.56 total mass and so of this, copper = 63.55/95.56=66.4%. So to get 10 grams of copper, use the formula 10g=66.4%xCuS so CuS=10/0.664=15.06 grams of CuS.
