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cluponka [151]
3 years ago
14

Increasing the temperature increases the vaporization rate of a liquid because the excess energy is used to break covalent bonds

Chemistry
1 answer:
igor_vitrenko [27]3 years ago
4 0

Answer:

False

Explanation:

False.  The molecules of liquid are hold in the liquid state due to intermolecular forces or Van de Waals forces , without affecting the molecule itself and its atomic bonds (covalent bonds).  When the temperature increases the kinetic energy of the molecules is higher , therefore they have more possibilities to escape from the attractive intermolecular forces and go to the gas state.

Note however that this is caused because the intermolecular forces are really weak compared to covalent bonds, therefore is easier to break the first one first and go to the gas state before any covalent bond breaks ( if it happens).

A temperature increase can increase vaporisation rate if any reaction is triggered that decomposes the liquid into more volatile compounds , but nevertheless, this effect is generally insignificant compared with the effect that temperature has in vaporisation due to Van der Waals forces.

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maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

<em>p₁</em> = <em>p</em>₁° - Δ<em>p</em> = 23.8 torr – 0.4430 torr = 23.4 torr

3 0
3 years ago
In general how are word equations written to describe chemical equations
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Which answer below correctly identifies the type of change and the explanation for the boiling of water?
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Explanation:

The boiling of water is a physical change because the original properties of the water is preserved.

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During the boiling of water, intermolecular bonds called hydrogen bonds between the water molecules are broken. This makes the individual molecules free.

Learn more:

Hydrogen bonds brainly.com/question/10602513

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