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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
Answer:
answer
<u>Cold Front - a zone separating two air masses, of which the cooler, denser mass is advancing and replacing the warmer. </u>
<u>Warm Front - a transition zone between a mass of warm air and the cold air it is replacing. </u>
<u>Stationary Front - a front between warm and cold air masses that is moving very slowly or not at all.</u>
Some characteristics of Ionic compounds by Mimiwhatsup: brittle, high melting point, conducts electricity when molten or dissolved in water.
