Answer: The heat needed to be removed to freeze 45.0 g of water at 0.0 °C is 15.01 KJ.
Explanation:
- Firstly, we need to define the term <em>"latent heat"</em> which is the amount of energy required "absorbed or removed" to change the phase "physical state; solid, liquid and vapor" without changing the temperature.
- Types of latent heat: depends on the phases that the change occur between them;
- Liquid → vapor, <em>latent heat of vaporization</em> and energy is absorbed.
- Vapor → liquid, latent heat of liquification and the energy is removed.
- Liquid → solid, <em>latent heat of solidification</em> and the energy is removed.
- Solid → liquid, <em>latent heat of fusion</em> and the energy is absorbed.
- In our problem, we deals with latent heat of freezing "solidification" of water.
- The latent heat of freezing of water, ΔHf, = 333.55 J/g; which means that the energy required to be removed to convert 1.0 g of water from liquid to solid "freezing" is 333.55 g at 0.0 °C.
- Then the amount of energy needed to be removed to freeze 45.0 g of water at 0.0 °C is (ΔHf x no. of grams of water) = (333.55 J/g)(45.0 g) = 15009.75 J = 15.01 KJ.
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Answer:
Physical Properties of Metals
:
Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).
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Answer:
Nonpolar covalent bond
Explanation:
when carbon atoms bonds to its self it form a nonpolar covalent bond
Two because it has six electrons in it's valence shell and it still has space for two more
