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.
Answer:
Any matter considered to be a fuel contains chemical energy
Energy has to be provided to initiate the reaction.
So yes. It is
Hey there!
I attached a screenshot of my chemistry notes from this year on equilibrium constants in particular. Feel free to reference this in the future if you wish.
Based on the information in my attachment, you could construct your constant like this:
I don't think it matters which order your reactants or products are in as long as you have them on the top and bottom where they need to be, so your answer will be B.
Hope this helped you out! :-)
Answer:
Explanation:
Given that:
Half life = 30 min
Where, k is rate constant
So,
The rate constant, k = 0.0231 min⁻¹
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given that:
The rate constant, k = 0.0231 min⁻¹
Initial concentration = 7.50 mg
Final concentration = 0.25 mg
Time = ?
Applying in the above equation, we get that:-
Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl