Answer:
![0.0397\ \text{mol}](https://tex.z-dn.net/?f=0.0397%5C%20%5Ctext%7Bmol%7D)
Explanation:
Number of molecules of water = ![2.39\times 10^{22}](https://tex.z-dn.net/?f=2.39%5Ctimes%2010%5E%7B22%7D)
= Avogadro's number = ![6.022\times 10^{23}\ \text{mol}^{-1}](https://tex.z-dn.net/?f=6.022%5Ctimes%2010%5E%7B23%7D%5C%20%5Ctext%7Bmol%7D%5E%7B-1%7D)
Number of moles is given by
![\dfrac{2.39\times 10^{22}}{6.022\times 10^{23}}](https://tex.z-dn.net/?f=%5Cdfrac%7B2.39%5Ctimes%2010%5E%7B22%7D%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%7D)
![=0.0397\ \text{mol}](https://tex.z-dn.net/?f=%3D0.0397%5C%20%5Ctext%7Bmol%7D)
The number of moles of water is
.
Can you show me the atoms please? I would be able to help.
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.
![\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20i%20%5Ctimes%20Kf%20%5Ctimes%20m%20%3D%20%20%201%20%5Ctimes%206.90%20%5C%C2%B0C%2Fm%20%20%5Ctimes%201.324m%20%3D%209.14%20%20%5C%C2%B0C)
where,
- i: van 't Hoff factor (1 for non-electrolytes)
- Kf: cryoscopic constant
- m: molality
The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:
![T = 80.26 \° C - 9.14 \° C = 71.12 \° C](https://tex.z-dn.net/?f=T%20%3D%2080.26%20%5C%C2%B0%20C%20-%209.14%20%5C%C2%B0%20C%20%3D%2071.12%20%5C%C2%B0%20C)
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
Learn more: brainly.com/question/2292439
A physical property is what a substance is like; it's directly observable. On the other hand, a chemical property is how a substance behaves; its reactivity.
Examples of a physical property are: color, texture, boiling point, freezing point, and melting point.
Examples of a chemical property are: flammability, combustion, and formation of a precipitate.
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