When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
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Here is the formula hope it helps.
The pH of the diluted HCl solution is 1.3.
Explanation:
Given:
The concentrated HCl solution of 8.0 M. The 1.5 mL of 8.0 M HCl is diluted with water to 250 mL volume.
To find:
The pH of the diluted HCl solution.
Solution
- The concentration of the HCl solution before dilution =

- The volume of the HCl solution taken for dilution =

- The concentration of the HCl solution after dilution =

- The volume of the HCl solution after dilution =

Using the Dilution equation:

The concentration of diluted HCl solution = 0.048 M

In the 1 M solution of HCl, there are 1 M of hydrogen ion, then the concentration of hydrogen ions in 0.048 M of HCl will be:
![[H^+]=1\times 0.048M=0.048 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1%5Ctimes%200.048M%3D0.048%20M)
The pH of the diluted HCl solution :
![pH=-\log [H^+]\\=-\log [0.048M]=1.18 \approx 1.3](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C%3D-%5Clog%20%5B0.048M%5D%3D1.18%20%5Capprox%201.3)
The pH of the diluted HCl solution is 1.3.
Learn more about the dilution equation here:
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Answer:
The loaded ship will weigh more. Thus it will sink down farther in the water in order to displace that much more water so that it continues to float.