The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the 
of the reaction, we use the equation:
Hence, the standard electrode potential of the cell is 4.53 V.
<span>Double displacement reaction is your answer.</span>
Answer:
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