Given the following data for water:
Heat of fusion = 334 J/g
Heat of vaporization = 2,256 J/g
Specific heat of solid = 2.09 J/g °C)
Specific heat of liquid = 4.184 J/g °C)
Specific heat of gas = 1.84 J/g °C)
Calculate how much energy is needed to change 100.0 grams of liquid water at 15.0 °C to vapor at 125.0 °C. (3 points)
Oa
O
b
44,000 J
89,400 J
104,000 J
266,000 J

Answers

Answer 1

The process of changing 100.0 grams of liquid water at 15.0 °C to vapor at 125.0 °C involves several steps, and we need to calculate the energy required for each step and then add them up:

1. Heating the liquid water from 15.0 °C to 100.0 °C:

q = m * Cp * ΔT
= 100.0 g * 4.184 J/g °C * (100.0 °C - 15.0 °C)
= 34,972 J

2. Vaporizing the liquid water at 100.0 °C:

q = m * Hvap
= 100.0 g * 2,256 J/g
= 225,600 J

3. Heating the water vapor from 100.0 °C to 125.0 °C:

q = m * Cp * ΔT
= 100.0 g * 1.84 J/g °C * (125.0 °C - 100.0 °C)
= 4,600 J

The total energy required is the sum of the three steps:

Q = q1 + q2 + q3
= 34,972 J + 225,600 J + 4,600 J
= 265,172 J

Therefore, the energy needed to change 100.0 grams of liquid water at 15.0 °C to vapor at 125.0 °C is approximately 265,172 J, which is closest to option (d) 266,000 J.

Related Questions

A flask filled to the 25.0 ml mark contain 29.97 g of a concentrated salt water solution. What is the density of the solution?

Answers

A concentrated saltwater solution weighing 29.97 g and fitting into a flask to the mark of 25.0 ml has a density of about 1199.2 g/L.

How is the density of the solution determined?

By dividing the solution's mass by its volume, we may get its density: density = mass/volume

We need to know the density of water at the solution's temperature as well as the capacity of the flask up to the 25.0 ml level in order to calculate the volume of the solution.

Since 1 mL = 0.001 L, volume is equal to 25.0 mL, or 0.0250 L.

Now, we may determine the solution's density as follows:

1199.2 g/L or 29.97 g/0.0250 L is what is referred to as density.

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CaCO3 + 2HCI =CaCl2 + H₂O + CO2
5. Calcium carbonate (CaCO3) combines with HCl to produce calcium chloride (CaCl₂),
water, and carbon dioxide gas (CO₂). How many grams of HCI are required to react with
6.35 mol CaCO3?

Answers

463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.

What is meant by molar mass?

Mass of one mole of substance is referred to as the molar mass. The molar mass of a substance can be calculated by adding up the atomic masses of all the atoms in a molecule.

Balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

6.35 mol CaCO₃ * 2 mol HCl / 1 mol CaCO₃ = 12.7 mol HCl

Now, we use the molar mass of HCl (36.46 g/mol) to convert from moles to grams: 12.7 mol HCl * 36.46 g/mol = 463.5 g HCl

Therefore, 463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.

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The two possible units of molarity are

Answers

Answer: The units for molarity are moles/liter.

Similarly, the equation to find molarity is moles divided by liters.

Explanation:

mol / L is a unit of molar concentration. These are the number of moles of dissolved material per liter of solution. 1 mol / L is also called 1M or 1molar. Mol / m3 is also a unit of molar concentration.

Molarity is expressed in units of moles per liter (mol / L). This is a very common unit, so it has its own symbol, which is the uppercase M. A solution with a concentration of 5 mmol / l is called a 5 M solution or has a concentration value of 5 mol.

The molar concentration of the solution is equal to the number of moles of the solute divided by the mass of the solvent (kilogram), and the molar concentration of the solution is equal to the number of moles of the solute divided by the volume of the solution (liter). increase.

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