Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Answer:522.93 centimeters
Explanation:
I calculated the numbers
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
Answer:
Joe Mama Formula
Explanation:
JoeMamic Acid is the answer
