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Viktor [21]
3 years ago
7

The heat of vaporization of a liquid is 84.0 J/g. How many joules of heat would it take to completely vaporize 172 g of this liq

uid at its boiling point?
Chemistry
1 answer:
Aliun [14]3 years ago
5 0

Answer:

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

Explanation:

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.

During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.

So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.

In this case, being:

  • Q=?
  • m= 172 g
  • L= 84 \frac{J}{g}

and replacing in the expression Q = m*L you get:

Q=172 g*84 \frac{J}{g}

Q=14,448 J

<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>

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Answer:

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Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

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M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

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0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

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Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

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Now we can calculate the concentration:

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