A material source utilized in research, as diaries or manuscripts. Hope this is the answer
Answer is: 20 ions, left side.
Unbalanced half reaction: C₅O₅²⁻(g) → CO₃²⁻(aq).
1) There are 5 carbon atoms on the left side of half reaction and 1 carbon atom on right, so first add coefficient 5 in fron of carbon dioxide to balance carbon atoms: C₅O₅²⁻(g) → 5CO₃²⁻(aq).
2) Because there are more oxygen atoms on the right, add OH⁻ ions on the left side of half reaction and water on the right: OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + H₂O(l).
3) Balance oxygen (25 atoms on boths side) and hydrogen (20 atoms on both side of half reaction) atoms:
Balanced half reaction: 20OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + 10H₂O(l).
Answer:
a) The temperature of the beaker rises as this transfer of heat goes on.
b) Check Explanation.
Explanation:
a) The heat lost by the piece of metal is normally gained by the all the components that it comes in contact with after the heating procedure.
(Heat lost by piece of metal) = (Heat gained by the cold water) + (Heat gained by the beaker).
So, since heat is also gained by the Beaker, its temperature should rise under normal conditions.
That is essentially what the zeroth law of thermodynamics about thermal equilibrium talks about.
If two bodies are at thermal equilibrium with reach other and body 2 is in thermal equilibrium with a third body, then body 1 and body 3 are also in thermal equilibrium
Temperature of the piece of metal decreases, temperature of water rises and the temperature of the beaker rises as they all try to attain thermal equilibrium.
b) In calorimetry, the aim is usually for the water (in this case) to take up all of the heat supplied by the piece of metal. Hence, the calorimeter is usually heavily insulated (or properly called lagged). Thereby, reducing the amount of heat that the calorimeter would gain.
But in cases where the heat lost to the insulated calorimeter isn't negligible, the heat capacity of the calorimeter is usually obtained and included it is included in the heat transfer calculations.
Hope this Helps!!!
Answer:
C) 1.15 × 10⁻⁷ mm
Explanation:
Step 1: Given data
Average distance between nitrogen and oxygen atoms: 115 pm
Step 2: Convert the distance to meters (SI base unit)
We will use the conversion factor 1 m = 10¹² pm.
115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m
Step 3: Convert the distance to millimeters
We will use the conversion factor 1 m = 10³ mm.
1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm
