when heat gained = heat lost
when AL is lost heat and water gain heat
∴ (M*C*ΔT)AL = (M*C*ΔT) water
when M(Al) is the mass of Al= 225g
C(Al) is the specific heat of Al = 0.9
ΔT(Al) = (125.5 - Tf)
and Mw is mass of water = 500g
Cw is the specific heat of water = 4.81
ΔT = (Tf - 22.5)
so by substitution:
∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)
∴Tf = 30.5 °C
Technically, the answer is iron. Oxygen has a melting point way below zero (-219 degrees celsius), ice becomes water AT room temperature and bromine is already a liquid at room temperature. So, iron has a melting point greater than room temperature due to the fact that metals are made up of giant structures of atoms in a regular arrangement, and there are strong forces of electrostatic attraction between positive metal ions and negative electrons, meaning that a lot of heat energy is required to break the bonds, i.e. a very high melting point, approx. 1500 degrees celsius. Hope this helps.
If you don't practice enough it's obviously going to be hard but if you practice enough it's going to be a piece of cake so don't think if it's going to be hard or not just think it's going to be worth the try at the very end
Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
