Answers are:
1) The balanced oxidation half reaction: 2I⁻(aq) → I₂(s) + 2e⁻.
Iodine is oxidized (lost electrons) from -1 to neutral charge (0).
2) The balanced reduction half-reaction: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻.
Hydrogen is reduced (gain electrons) from +1 to neutral charge.
3) The oxidation <span>reaction takes place at the anode.</span>
the room will either cool down or warm up the water until it is the same temperature as the room around it.
Polysaccharides, that allow sugar to polymerize
Answer:
well physical weathering is the process of a element or acts of water rock or other items that take its toll on items
Explanation:
so i would,WITH CONFIDENCE C.
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
