Answer : The final temperature of the metal block is, 
Explanation :
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of aluminum = 55 g
= mass of water = 0.48 g
= final temperature = ?
= temperature of aluminum =
= temperature of water =
= specific heat of aluminum = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]](https://tex.z-dn.net/?f=55g%5Ctimes%200.900J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%3D-%5B0.48g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%5D)
Thus, the final temperature of the metal block is,
The ice sculpture would take longer to melt than the ice cube
You that Mg is the metal if in the periodic table it lands between group 1-12. So yeas Mg is a metal because it lands in group 2.
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
