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Hitman42 [59]
3 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]3 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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Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

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  • pKa =?

pKa = –Log Ka

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<h3>How to determine the molarity of HCl </h3>
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pH = pKa + Log[salt]/[acid]

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Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>
  • Molarity of HCl = 0.0069 M
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Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

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