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.
Answers
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.
Answer:
-21.03 kJ/mol
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.
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Answers
The final temperature of the water, T2 = 38.57°C
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Given the following data:
Quantity of energy = 420JMass = 35 gramsInitial temperature, T1 = 10°CThe specific heat capacity of water is 4.2 J/g°C.
To find the final temperature of the water (T2):
Mathematically, the quantity of energy (heat capacity) is given by the formula;
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Q represents the heat capacity or quantity of heat.M represents the mass of an object.C represents the specific heat capacity of water.dt represents the change in temperature.Substituting the values into the formula, we have;
[tex]420 = 3.5 \; * \; 4.2 \; * \; dt[/tex]
[tex]420 = 14.7 \; * \; dt\\\\dt = \frac{420}{14.7}[/tex]
Change in temperature, dt = 28.57°C
Next, we would solve for the final temperature by using this formula;
[tex]dt = T2 - T1[/tex]
[tex]28.57 = T_{2} - 10\\\\T_{2} = 28.57 \; + \; 10\\\\T_{2} = 38.57[/tex]
Final temperature, T2 = 38.57°C
Therefore, the final temperature of the water, T2 is equal to 38.57°C
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Explanation:
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Considering the definition of enthalpy of reaction, the correct answer is option C. a reaction with a negative enthalpy indicates that the reaction is exothermic.
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2. 39.2
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Using your answers from part C, complete the chemical formula of each compound. If the number of ions is 1, leave that
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Drag each number to the correct location. Numbers may be used more than once
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Answer:
1;1, 2;1, 3;1, 1;2, 1;1, 3;2, 1;3, 2;3, 1;1.
I hope this will help.
Answer:
This is what the answer looks like
Explanation:
The process of diffusion can be described as: Movement of water from the area of high concentration of solute to the area of low concentration of solute. Movement of solute from the area of low concentration of solute to the area of high concentration of solute. Movement of water from the area of low concentration of solute to the area of high concentration of solute. Movement of solute from the area of high concentration of solute to the area of low concentration of solute.
Answers
Answer:
Movement of water from the area of high concentration of solute to the area of low concentration of solute.
Explanation:
Diffusion can be defined as the movement of gas molecules from a region of high concentration to a region of lower concentration until an equilibrium is attained or reached. This mixing of gas molecules in the liquid or gaseous state is due to their kinetic energy of random motion i.e without requiring bulk motion.
This ultimately implies that, diffusion is a phenomenon through which gas molecules in either a gaseous or liquid state spread out in response to a concentration gradient, which is simply the difference in concentration of the gases across a space.
During diffusion, molecules move from areas of high concentration to areas of low concentration. Also, diffusion does not require energy to move molecules.
Hence, the process of diffusion can be described as movement of water from the area of high concentration of solute to the area of low concentration of solute.
A solute can be defined as a substance that is typically dissolved in a liquid solvent such as water, coffee, tea, etc., to produce a solution.
