Answer:
(a) from the anode to the cathode through the external circuit
Explanation:
In an electrochemical cell, there are two half cells; the oxidation half cell and reduction half cell. Oxidation typically refers to loss of electrons and reduction refers to gain of electrons.
Electrons always flow from the anode to the cathode or from the oxidation half cell to the reduction half cell.
The electrical circuit in an electrochemical cell confirms the flow of electron. Usually a light bulb is attached. The correct option is;
(a) from the anode to the cathode through the external circuit
Because the pot isn’t water it just gets really hot and you can burn yourself if you touch it
A. SO2Cl2(g) --> SO2(g) + Cl2(g)
<span>1 mole of SOCl2 becomes 1 mole SO2 and 1 mole Cl2 </span>
<span>1 mole --> 2 moles </span>
<span>entropy increases </span>
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
