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Chapter 1: Problem 7

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)

Short Answer

Expert verified
(A) is the correct choice as it represents the formula for calculating the heat gained by water: \(q = mc\Delta T\), correctly substituting the values given in the question.

Step by step solution

01

Clarify the Given Values

Given values are, mass of water = \(25.0 \) mL; since we assume the density of water as \(1.00 \) g/mL, the mass of water is also \(25.0 \) g. The specific heat capacity of the substance = \(4.18 \) J/(g掳C) and \(\Delta T = 2.2^{\circ}\mathrm{C}\). The formula for energy lost or gained is \(q = mc\Delta T\).
02

Use Given Values with the Formula

By substituting the above given values into the formula \(q = (25.0 \mathrm{g})(4.18 \mathrm{J / g掳C})(2.2^{\circ}\mathrm{C})\). In this equation, the weight of water needs to be used, but not the weight of the solution (which is the combination of water and NaNO3). Hence, the weight of NaNO3 or the density of the solution should not affect the calculation of heat gained by water.
03

Confirm the Correct Option

Now, it is clear that expression (A) \((25.0)(4.18)(2.2)\) exactly matches our equation, and hence, it's the correct option.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Heat Capacity
Heat capacity is a crucial concept in thermodynamics that describes how much heat energy is needed to raise the temperature of a substance by a certain amount. It's expressed in units such as Joules per gram per degree Celsius (J/g掳C). In the context of our exercise, the specific heat capacity of water is given as 4.18 J/g掳C.
  • This value tells us how much energy it takes to raise the temperature of 1 gram of water by 1掳C.
  • Knowing the heat capacity helps us calculate the total heat gained or lost when a substance undergoes a temperature change.
When salt like sodium nitrate (\(\mathrm{NaNO}_{3}\)) is dissolved in water, we can use the heat capacity to compute the change in thermal energy. This understanding helps explain how heat is transferred during the dissolution process.
Exploring the Dissolution Process
The dissolution process involves dissolving a solute in a solvent, forming a solution. In our problem, 1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved in 25.0 mL of water. This process often involves energy changes.
  • Endothermic processes absorb heat, while exothermic processes release heat.
  • For sodium nitrate, dissolving typically causes the solution's temperature to change, indicating an exchange of heat energy.
When you calculate the heat involved, you focus on the heat exchange with the solvent (water, in this case) rather than the entire solution.
Tracking Temperature Change
The temperature change (\(\Delta T\)) is a direct sign of energy change in a system. In this problem, the temperature of the water increases by 2.2掳C, signaling an energy gain.
  • Temperature change is essential for calculating heat energy using the formula \(q = mc\Delta T\), where \(q\) is the heat energy.
  • By knowing both the temperature change and the specific heat capacity of water, we can determine how much heat is absorbed or released.
This temperature increase hints at an exothermic dissolution, as energy from the system is transferred to the surroundings.
Examining Calorimetry
Calorimetry is the science of measuring heat changes in chemical reactions. In this exercise, we use calorimetry principles to measure the heat gained by water when sodium nitrate dissolves. This involves working with the calorimeter, a device that captures changes in thermal energy.
  • The heat gained or lost by water can be calculated with \(q = mc\Delta T\)
  • Calorimetry ensures we focus only on the system of interest, excluding external factors like the container's heat capacity.
Through calorimetry, you can effectively determine the energy changes associated with physical and chemical processes, providing insights into the system's behavior.

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Most popular questions from this chapter

A 22.0 gram sample of an unknown gas occupies 11.2 liters at standard temperature and pressure. Which of the following could be the identity of the gas? (A) \(\mathrm{CO}_{2}\) (B) \(\mathrm{SO}_{3}\) (C) \(\mathrm{O}_{2}\) (D) He

Even though it is a noble gas, xenon is known to form bonds with other elements. Which element from the options below would xenon most likely be able to bond with? (A) Lithium (B) Argon (C) Fluorine (D) Carbon

Which of the following pairs of elements is most likely to create an interstitial alloy? (A) Titanium and copper (B) Aluminum and lead (C) Silver and tin (D) Magnesium and calcium

The ionization energies for an element are listed in the table below. \(\begin{array}{lllll}{\text { First }} & {\text { Second }} & {\text { Third }} & {\text { Fourth }} & {\text { Fifth }} \\ {8 \mathrm{eV}} & {15 \mathrm{eV}} & {80 \mathrm{eV}} & {109 \mathrm{eV}} & {141 \mathrm{eV}}\end{array}\) Based on the ionization energy table, the element is most likely to be (A) Sodium (B) Magnesium (C) Aluminum (D) Silicon

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which net ionic equation below represents a possible reaction that takes place when a strip of magnesium metal is oxidized by a solution of chromium (III) nitrate? (A) \(\operatorname{Mg}(s)+\operatorname{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)+3 \mathrm{NO}_{3}^{-}(a q)\) (B) \(3 \mathrm{Mg}(s)+2 \mathrm{Cr}^{3+} \rightarrow 3 \mathrm{Mg}^{2+}+2 \mathrm{Cr}(s)\) (C) \(\mathrm{Mg}(s)+\mathrm{Cr}^{3+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Cr}(s)\) (D) \(3 \mathrm{Mg}(s)+2 \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow 3 \mathrm{Mg}^{2+}(a q)+2 \mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q)\)
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