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In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g h2o initially at 23.2°c. the final temperature of the solution was 31.8°c. what was the change in enthalpy for the dissolution of this compound? give your answer in units of joules per gram of compound. assume that the specific heat of the solution is the same as that of pure water, 4.18 j ⁄ g ⋅ °c.
A) 2.7 x 10^2 J/g
B) -3.1 x 10^3 J/g
C) -3.1 x 10^2 J/g
D) 3.1 x 10^2 J/g

Respuesta :

Given:

Mass of the ionic compound  = 10.00 g

Mass of water = 75.0 g

Initial temperature of water T1= 23.2 C

Final temperature of water T2 = 31.8 C

Specific heat of water c = 4.18 J/gC

To determine:

Enthalpy of dissolution of the ionic compound

Explanation:

Heat gained by water is given as:

Q = mcΔT

m = mass of water

c = specific heat

ΔT = change in temp of water = T2-T1

Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J

Now,

Heat gained by water = heat lost by the ionic compound (i.e. heat of dissolution)

Thus, q(ionic) = 2696 J

ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g

Ans: A) enthalpy change = 2.7*10² J/g



soneyedesola44

The change in enthalpy for dissolution of this compound is 2.70 x 10²J/g

  • The mass of the soluble ionic compound = 10.00g
  • The mass of water = 75.0g
  • The initial temperature of water = 23.2⁰C (t1)
  • The final temperature of water = 31.8⁰C (T2)
  • Specific heat of water c = 4.18 J/g.⁰C

Further Explanation

Step 1:

To determine the heat gain by the water, you have use the formula for heat transfer. The heat content (Q) which is the amount of heat lost or gain depends on the specific heat of the substances, (c) and its mass, (m). Therefore, this can express as Heat transfer = (Mass) (specific heat) (temperature change)

Q = mcΔT

  • Q is the heat content
  • M is the mass of water  
  • ΔT is the change in temperature (T2-T1)

If you substitute the values, you have:

Q = (75.0 g) x (4.18 J/g.⁰C) x (31.8⁰C-23.2⁰C)

75.0 g x 4.18 J/g.⁰C x (8.6⁰C)  

= 2696 J

Step 2:

The heat gain by water should be equals to the amount of heat lost by the soluble ionic compound and can express as:

q (h20) = q (ionic) = 2696 J

q (Ionic) = m x ΔH

ΔH = q (Ionic)/m

= 2696 J/10.00g

= 270J/g

= 2.70 x 10²J/g

Thus, the change in enthalpy for dissolution of this compound is 2.70 x 10²J/g

LEARN MORE:

  • In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g h2o initially at 23.2°c https://brainly.com/question/9362336
  • In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g h2o initially at 23.2°c https://brainly.com/question/11509593

KEYWORDS:

  • heat transfer
  • soluble ionic solution
  • change in enthalpy
  • units of joule per gram
  • specific heat

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