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REY [17]
2 years ago
6

Of the following statements about water's latent heat and changes in state, which is/are true? choose all that apply.

Chemistry
1 answer:
postnew [5]2 years ago
4 0

When water evaporates, its latent heat of evaporation absorbs heat.

<h3>What is Latent heat?</h3>

A body or thermodynamic system can release or absorb latent heat during a constant-temperature process, often a first-order phase transition.

There are two types of latent heat:

  1. The heat of freezing
  2. The heat of vaporization

The statements about water's latent heat and changes in state are:

  • The heat of freezing is the quantity of thermal energy produced as a liquid freezes, and the heat of vaporization is the quantity of thermal energy needed to change a liquid into a gas.
  • Water may move energy from one location to other thanks to latent heat.
  • There aren't any weak (hydrogen) bonds between water molecules in the vapor state.
  • Hurricanes are propelled by the latent heat of condensation, which occurs when water vapor condenses and turns into a liquid.
  • All water molecules in the solid state are connected by flimsy (hydrogen) bonds.

To learn more about latent heat, refer:

brainly.com/question/5401454

#SPJ4

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Answer:

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Explanation:

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3 years ago
Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 g of naphthalene is burned in a bomb calorim
Serggg [28]

Answer:

\Delta E_{rxn} for combustion of naphthalene is -5164 kJ/mol

Explanation:

\Delta E_{rxn}=C_{calorimeter}\times \Delta T_{calorimeter}

where, C refers heat capacity and \Delta T refers change in temperature.

Here, \Delta T_{calorimeter}=(32.33-24.25)^{0}\textrm{C}=8.08^{0}\textrm{C}

So, \Delta E_{rxn}=(5.11\frac{kJ}{^{0}\textrm{C}})\times (8.08^{0}\textrm{C})=41.3kJ

\Delta E_{rxn} is generally expressed in terms of per mole unit of reactant. Also, \Delta E_{rxn} should be negative as it is an exothermic reaction (temperature increases).

Molar mass of naphthalene is 128.17 g/mol

So, 1.025 g of naphthalene = \frac{1.025}{128.17}moles of naphthalene

                                              = 0.007997 moles of naphthalene

\Delta E_{rxn}=-\frac{41.3kJ}{0.007997mol}=-5164kJ/mol

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