When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
Answer:
3NaOH + H3PO4 ---> 3H2O + Na3PO4
Explanation:
15.9643 gramns are in 2.3 moles of lithium
Answer:
We have to take 37.5 mL of a 0.400 M solution
Explanation:
Step 1: Data given
Stock volume = 100 mL = 0.100L
Stock concentration 0.400 M
Volume of solution he wants to make = 100 mL = 0.100L
Concentration of solution he wants to make = 0.150 M
Step 2: Calculate the volume of 0.400 M CuSO4 needed
C1*V1 = C2*V2
⇒with C1 = the stock concentration = 0.400M
⇒with V1 = the volume of the stock = TO BE DETERMINED
⇒with C2 = the concentration of the solution he wants to make = 0.150 M
⇒with V2 = the volume of the solution made = 0.100 L
0.400 M * V1 = 0.150M * 0.100L
V1 = (0.150M*0.100L) / 0.400 M
V1 = 0.0375 L = 37.5 mL
We have to take 37.5 mL of a 0.400 M solution
