Chemistry, 15.07.2019 02:30 maggie123456751

How much heat is required to vaporize 33.8 g of water at 100 ∘c?

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Answer from: AquaNerd5706

The heat required to vaporize 33.8 g of water at ${\text{100 }}^\circ{\text{C}}$ is $\boxed{76.353{\text{2 kJ}}}$ .

Further explanation:

Properties are categorized into two types:

1. Intensive properties:

These properties depend on the nature of the substance and not on the size of the system. If the system is further divided into a number of subsystems, the values of intensive properties remain unchanged. Temperature, refractive index, molarity, concentration, pressure, and density are some of the examples of intensive properties.

2. Extensive properties:

These properties depend on the amount of the substance. These are additive in nature if a single system is divided into several subsystems. Mass, heat, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.

Heat of vaporization:

It is the amount of heat that is needed to convert a unit mass of the liquid into its vapors. It is also known as the enthalpy of vaporization or heat of evaporation. It is represented by $\Delta {H_{{\text{vap}}}}$ . Its unit is kJ/mol or J/mol.

The formula to calculate the moles of water is as follows:

${\text{Moles of water}}=\frac{{{\text{Given mass of water}}}}{{{\text{Molar mass of water}}}}$ …… (1)

The given mass of water is 33.8 g.

The molar mass of water is 18.01 g/mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Moles of water}}&=\left({{\text{33}}{\text{.8 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{18}}{\text{.01 g}}}}}\right)\\&=1.87{\text{6 mol}}\\\end{aligned}$

The heat of vaporization of water at ${\text{100 }}^\circ{\text{C}}$ is 40.7 kJ/mol. This implies 40.7 kJ of energy is required to convert 1 mole of water into vapors. So the heat required to convert 1.876 moles of water is calculated as follows:

$\begin{aligned}{\text{Heat required}}&=\left( {{\text{1}}{\text{.876 mol}}}\right)\left({\frac{{{\text{40}}{\text{.7 kJ}}}}{{{\text{1 mol}}}}}\right)\\&=76.353{\text{2 kJ}}\\\end{aligned}$

Therefore the heat required to vaporize 33.8 g of water at ${\text{100 }}^\circ{\text{C}}$ is 76.3532 kJ.

Learn more:

1. Calculate the enthalpy change using Hess’s Law:

2. Find the enthalpy of decomposition of 1 mole of MgO:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: heat of vaporization, enthalpy of vaporization, heat of evaporation, 33.8 g, water, 40.7 kJ/mol, 76.3532 kJ, heat, intensive, extensive, molar mass of water, 18.01 g/mol, 1.876 mol.

Answer from: Quest

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

Answer from: Quest

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

$^{m}_{4}x$ and $^{m}_4y$

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

$^{m}_{5}x^{-}$

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be $^{9}_{5}be^{-}$ and $^{9}_{4}be$ .

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.

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