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Hitman42 [59]
2 years ago
15

The specific heat of liquid bromine is 0.226 J/g-K. How much heat (J) is required to raise the temperature of 10.0 mL of bromine

from 25.00 °C to 27.30 °C? The density of liquid bromine: 3.12 g/mL
Chemistry
1 answer:
Viefleur [7K]2 years ago
7 0

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

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Answer:

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1.0 mol of O₂ produce → 2.0 mol of H₂O, from stichiometry.

0.1875 mol of O₂ produce → ??? mol of H₂O.

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1.0 mol of H₂O represents → 22.4 L, at STP.

3.75 mol of H₂O represents → ??? L.

<em>∴ The no. of liters of water vapor will be produced </em>= (3.75 mol)(22.4 L)/(1.0 mol) = <em>8.4 L.</em>

<em></em>

<em>So, the right choice is: D)  8.40 L H₂O(g).</em>

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2 years ago
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