Answer:
3.3765 Mol O2
Explanation:
There is no work for this problem
Answer:
The particles of the medium just vibrate in place.
Explanation:
As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don't actually travel along with the wave.
Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g
Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution: Molarity (M) = moles solute. liters solution.
Answer:
∇T = 51.68°C
Explanation:
Mass = 150g
Heat Energy (Q) = 1.0*10³J
Change in temperature ∇T = ?
Q = mc∇T
Q = heat energy
M = mass
C = specific heat capacity of the gold = 0.129j/g°C
∇T = change in temperature
Q = Mc∇T
1.0*10³ = 150 * 0.129 * ∇T
1000 = 19.35∇T
Solve for ∇T
∇T = 1000 / 19.35
∇T = 51.679°C = 51.68°C
The change in temperature of gold was 51.68°C
