Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 46

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. What is the approximate amount of heat released during the reaction? \(\begin{array}{ll}{\text { (A) }} & {1.5 \mathrm{kJ}} \\ {\text { (B) }} & {2.9 \mathrm{kJ}} \\ {\text { (C) }} & {5.9 \mathrm{kJ}} \\ {\text { (D) }} & {11.8 \mathrm{kJ}}\end{array}\)

Short Answer

Expert verified
The approximate amount of heat released during the reaction is 5.88kJ. Hence, the closest answer is (C) 5.9 kJ.

Step by step solution

01

Understand the Concept

In a calorimeter, the heat released or absorbed during a chemical reaction is measured. Here, given the temperature change and the density and specific heat of the solution, the heat released can be calculated using q=mc\(\Delta\)T where q is heat, m is mass, c is specific heat capacity and \(\Delta\)T is the change in temperature.
02

Determine the Parameters

The volume of the mixture in the calorimeter is 100.0 mL of HCN + 100.0 mL NaOH = 200.0 mL. Since the density is identical to water, we consider 1 mL = 1 g. So, the mass(m) of the mixture is 200.0 g. The specific heat capacity(c) is provided as 4.2 J/g°C. The change in temperature(\(\Delta\)T) is the final temperature(28.5°C) - initial temperature(21.5°C) = 7.0°C.
03

Calculate Heat q

Substitute the parameters we have determined into the formula: q = mc\(\Delta\)T. The calculation will then look as follows: q = 200.0g * 4.2 J/g°C * 7.0°C = 5880 J.
04

Convert to Kilojoules

As our answer options are given in kilojoules, we need to convert joules to kilojoules by dividing by 1000: q = 5880J/1000 = 5.88 kJ.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ What will happen in the salt bridge as the reaction progresses? (A) The Na' ions will flow to the Cu/Cu' half-cell. (B) The Br' ions will flow to the Cu/Cu' half-cell. (C) Electrons will transfer from the Cu/Cu' half-cell to the Zn/Zn" half cell. (D) Electrons will transfer from the \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half- cell to the Cu/Cu' half cell.

\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_{2}+7 \mathrm{H}_{2} \mathrm{O}\) Which of the following statements about the reaction given above is NOT true? (A) The oxidation number of chromium changes from \(+6\) to \(+3 .\) (B) The oxidation number of iodine changes from \(-1\) to 0. (C) The oxidation number of hydrogen changes from +1 to 0. (D) The oxidation number of oxygen remains the same.

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

Which of the substances would be soluble in water? (A) Ethylene glycol only, because it has the longest bond lengths (B) Acetone only, because it is the most symmetrical (C) Ethanol and ethylene glycol only, because of their hydroxyl (-OH) (D) All three substances would be soluble in water due to their permanent dipoles.

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free