Use the formula
first step:
Use the formula
molarity= mole/liter
change ml to l
plug in data
to get .1=mole/.25 or .1M*.25liter
which =.025 moles
then divide .025 moles by two because there are two OH in Sr(OH)2
then multiply that by 265.76 (the molar mass of water)
.0125*265.76
which is 3.32grams this is your answer
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.
For
the reaction 2SO2 + O2 -> 2 SO3, we first determine which is the excess
reactant between SO2 and O2. We list down the molar mass of the reactants:
Molar
mass of SO2 = 64.0638 g/mol
Molar
mass of O2 = 32 g/mol
Using
the stoichiometry of the reaction, we then calculate the amount of oxygen that
will react with 90.0 g of SO2.
90.0
g SO2 x 1 mol SO2/64.0638 g x 1 mol O2/ 2 mol SO2 x 32 g O2/mol = 22.4776 g O2
<span>
</span>
<span>Thus,
we can conclude that O2 is the excess reactant while SO2 is the limiting
reactant. Subtracting 22.4776 g O2 from the initial 100.0 g O2, we get 77.5224
g O2 left after the complete reaction of 90.0 g SO2. </span>
Answer:
Hey there!
Antimony is the element with an atomic number of 51. The element is in group 15, and in period 5.
Let me know if this helps :)