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
Answer:
The answer is option 3.
Explanation:
Option 3 shows a balanced equation.
Answer: The potential of the following electrochemical cell is 1.08 V.
Explanation:
=-0.74V[/tex]
=0.34V[/tex]
The element with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials, when concentration is 1M.
![E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BCu%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Thus the potential of the following electrochemical cell is 1.08 V.
Hey there!
Great question;)
Answer:Physical change
Explanation: When sugar mixes with water, at the end, the chemical formulas are the same. Nothing has changed!
I hope this helps;)
Reactants would be the missing word in the sentence because they make up the products in a chemical reaction
