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TIMED HELP ASAP

19.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the dissolution of MgSO₄. (The specific heat of water is 4.184 J/g・°C and the density of the water is 1.00 g/mL). You can assume that the specific heat of the solution is the same as that of water.

User Osman Cea
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2 Answers

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Answer:

-21.03 kJ/mol

Step-by-step explanation:

∆H is enthalpy. Enthalpy is the total heat content of a system.

So we can establish that ∆H = q (heat)

The formula for heat (q)

q = cm∆T

c = specific heat capacity

m = mass of substance

∆T= change of temperature

Since we are calculating the enthalpy of the SOLUTION. We must account for both the mass of water and the mass of MgSO₄ in our q formula.

All you gotta do is plug and chug at this stage.

∆H = q = cm∆T = (4.184)(19.11+100.0)(6.70) = 3338.986808 Joules

We have now calculated the heat (aka enthalpy) of the solution.

BUT

Remember! The problem asked for enthalpy in kj....

Use this conversion factor.

1000 J = 1 kJ

3338.986808 Joules * 1kJ/1000 J = 3.338986808 kJ

We were asked to find ∆H for the dissolution of MgSO4 in units of kj/mol so we are not finished.

Take the grams of MgSO4 in the problem and convert it into moles using its molar mass.

MgSO4 molar mass = 120.3676 g

1 mol = 120.3676 g MgSO4 <---- Use this as a conversion factor

19.11 g MgSO4 * 1 mol MgSO4/120.3676 g = 0.15876365 mol MgSO4

Now that you've calculated the moles of MgSO4 in this solution. You can divide your heat by it.

q dissolution = 3.338.986808 kJ/ 0.15876365 mol = 21.03 kj/mol

Note that the problem tells us that the temperature of water increases.

This means that the water is experiencing an endothermic process (heat is being absorbed from MgSO4) . Mathematically, this would be indicated by a positive sign. ---> +q

We can assume MgSO4 is losing heat as it is placed into the water. It is experiencing an exothermic process (heat is being lost). Mathematically, this would be indicated by a negative sign. ----> -q

So if we're calculating the enthalpy for the dissolution of MgSO₄ ....the value we've arrived at must be negative.

Answer

∆H = 21.03 kj/mol :)))))

I hope that helped...I feel like my explanation was a bit convoluted.

TIMED HELP ASAP 19.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature-example-1
User Mackendy
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Answer:

TIMED HELP ASAP

19.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the dissolution of MgSO₄. (The specific heat of water is 4.184 J/g・°C and the density of the water is 1.00 g/mL). You can assume that the specific heat of the solution is the same as that of water.

User Shahzad
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4.2k points