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Bond [772]
3 years ago
13

So my science teacher says if you put water in a cup and then put oil in it the oil is "more dense than water" witch means the o

il will go to the bottom but I thought it would go to the top plz help so I can pass my quiz
Chemistry
2 answers:
Neporo4naja [7]3 years ago
7 0
The oil is heavier (aka "denser") than the water. It is more heavily packed with molecules than the water. The oil slowly seeps to the bottom and the water rises above it. The two do not immediately mix due to the insolubility of oil. I hope this helps and please don't forget to make me the Brainliest :-)
laila [671]3 years ago
6 0
The oil would sink, due to the density of it
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How much energy (heat) is required to convert 248 g of water from 0 oC to 154 oC? Assume that the water begins as a liquid, that
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Answer:

The total heat required is 691,026.36 J

Explanation:

Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L

Where Q: amount of heat, m: mass and L: latent heat

On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).

In this case, the total heat required is calculated as:

  • Q  for liquid water.  This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C

Q= c*m*ΔT

Q=4.184\frac{J}{g*C} *248 g* (100 -0 )C

Q=103,763.2 J

  • Q  for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then\frac{248 g}{18 \frac{g}{mol} } =13.78 moles )

Q= m*L

Q=13.78moles*40.79 \frac{kJ}{mol}

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)

  • Q for temperature change from  100.0 ∘ C  to  154 ∘ C, this is, the sensible heat of steam from 100 °C to 154°C.

Q= c*m*ΔT

Q=1.99\frac{J}{g*C} *248 g* (154 - 100 )C

Q=25,176.96 J

So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J

<u><em>The total heat required is 691,026.36 J</em></u>

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