Answer:
16.2 J
Explanation:
Step 1: Given data
- Specific heat of liquid bromine (c): 0.226 J/g.K
- Volume of bromine (V): 10.0 mL
- Initial temperature: 25.00 °C
- Final temperature: 27.30 °C
- Density of bromine (ρ): 3.12 g/mL
Step 2: Calculate the mass of bromine
The density is equal to the mass divided by the volume.
ρ = m/V
m = ρ × V
m = 3.12 g/mL × 10.0 mL
m = 31.2 g
Step 3: Calculate the change in the temperature (ΔT)
ΔT = 27.30 °C - 25.00 °C = 2.30 °C
The change in the temperature on the Celsius scale is equal to the change in the temperature on the Kelvin scale. Then, 2.30 °C = 2.30 K.
Step 4: Calculate the heat required (Q) to raise the temperature of the liquid bromine
We will use the following expression.
Q = c × m × ΔT
Q = 0.226 J/g.K × 31.2 g × 2.30 K
Q = 16.2 J
The dark side your welcome
<span>2.44 × 10–2 m by 1.4 × 10–3 m by 8.4 × 10–3 m
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2.9 x 10-7 m3
Answer:
Number of moles of CoCl3 = 0.09 moles
Explanation:
Given data:
Mass of CoCl₃ = 12.1 g
Number of moles = ?
Solution:
First of all we will calculate the molar mass.
Molar mass of CoCl3 = 28.01 + (35.5×3)
Molar mass of CoCl3 = 28.01 + 106.5
Molar mass of CoCl3 = 134.51 g/mol
Number of moles of CoCl3 = mass / molar mass
Number of moles of CoCl3 = 12.1 g/134.51 g/mol
Number of moles of CoCl3 = 0.09 moles
