Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 78

Write a balanced equation for the reaction that occurs in each (a) Calcium metal is heated in an atmoof the following cases: sphere of oxygen gas. (b) Copper oxide is heated in an atmosphere of hydrogen gas. (c) Chlorine reacts with nitrogen gas. (d) Boron tribromide reacts with water.

Short Answer

Expert verified
The balanced equations for the given reactions are: (a) 2Ca + O2 → 2CaO (b) CuO + H2 → Cu + H2O (c) 3Cl2 + N2 → 2NCl3 (d) BBr3 + 3H2O → H3BO3 + 3HBr

Step by step solution

01

(a) Identifying the reaction between calcium and oxygen

To determine the product of the reaction between calcium metal (Ca) and oxygen gas (O2), we need to know the valency of the calcium and oxygen. Calcium has a valency of +2 and oxygen has a valency of -2. Based on the valency, calcium and oxygen come together to form calcium oxide (CaO). Now we need to balance the equation: Ca + O2 → CaO.
02

(a) Balancing the equation

To balance the equation, we need to make sure that the number of atoms of each element is equal on both sides. Ca + O2 → CaO 1Ca + 2O → 1Ca + 1O Since there are 2 oxygen atoms on the left side but only 1 on the right side, multiply the right side by 2 to balance the equation: Ca + O2 → 2CaO Now, we need to balance the number of calcium atoms. Multiply the calcium atom on the left side of the equation by 2: 2Ca + O2 → 2CaO Now the equation is balanced.
03

(b) Identifying the reaction between copper oxide and hydrogen

To determine the product of the reaction between copper(II) oxide (CuO) and hydrogen gas (H2), we need to know that hydrogen is a reducing agent that will reduce CuO to elemental copper (Cu) and form water (H2O) as a byproduct: CuO + H2 → Cu + H2O.
04

(b) Balancing the equation

To balance the equation, we need to make sure that the number of atoms of each element is equal on both sides. CuO + H2 → Cu + H2O 1Cu + 1O + 2H → 1Cu + 2H + 1O The equation is already balanced.
05

(c) Identifying the reaction between chlorine and nitrogen

The reaction between chlorine gas (Cl2) and nitrogen gas (N2) produces nitrogen trichloride (NCl3). Now we need to balance the equation: Cl2 + N2 → NCl3.
06

(c) Balancing the equation

To balance the equation, we need to make sure the number of atoms of each element is equal on both sides. Cl2 + N2 → NCl3 2Cl + 2N → 1N + 3Cl To balance the number of nitrogen atoms, multiply the NCl3 on the right side by 2: Cl2 + N2 → 2NCl3 2Cl + 2N → 2N + 6Cl To balance the number of chlorine atoms, multiply the Cl2 on the left side by 3: 3Cl2 + N2 → 2NCl3 6Cl + 2N → 2N + 6Cl Now the equation is balanced.
07

(d) Identifying the reaction between boron tribromide and water

The reaction between boron tribromide (BBr3) and water (H2O) produces boric acid (H3BO3) and hydrobromic acid (HBr). Now we need to balance the equation: BBr3 + H2O → H3BO3 + HBr.
08

(d) Balancing the equation

To balance the equation, we need to make sure the number of atoms of each element is equal on both sides. BBr3 + H2O → H3BO3 + HBr 1B + 3Br + 2H + 1O → 1B + 1O + 3H + 1H + 1Br To balance the number of bromine atoms, multiply HBr on the right side by 3: BBr3 + H2O → H3BO3 + 3HBr 1B + 3Br + 2H + 1O → 1B + 1O + 3H + 3H + 3Br Now the equation is balanced.

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

Mercury in the environment can exist in oxidation states \(0,\) \(+1,\) and \(+2 .\) One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111 ), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy $\left(1 \mathrm{ev}=1.602 \times 10^{-19} \mathrm{~J}\right)$ The oxygen on the mineral surface gave emitted electron energies at $531 \mathrm{eV}\(, corresponding to the \)1 s$ orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for O. (a) Calculate the wavelength of the \(\mathrm{X}\) rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the \(1 s\) electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case?

Elements in group 17 in the periodic table are called the halogens; elements in group 16 are called the chalcogens. (a) What is the most common oxidation state of the chalcogens compared to the halogens? (b) For each of the following periodic properties, state whether the halogens or the chalcogens have larger values: atomic radii, ionic radii of the most common oxidation state, first ionization energy, second ionization energy.

Give three examples of +2 ions that have an electron configuration of $n d^{10}(n=3,4,5 \ldots)$

Write balanced equations for the following reactions: (a) boron trichloride with water, (b) cobalt (II) oxide with nitric acid, (c) phosphorus pentoxide with water, (d) carbon dioxide with aqueous barium hydroxide.

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. exo (Section 6.2 ) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength \(58.4 \mathrm{nm} .\) (a) What is the energy of a photon of this light, in joules? (b) Write an equation that shows the process corresponding to the first ionization energy of \(\mathrm{Hg}\). (c) The kinetic energy of the emitted electrons is measured to be \(1.72 \times 10^{-18} \mathrm{~J}\). What is the first ionization energy of \(\mathrm{Hg}\), in $\mathrm{kJ} / \mathrm{mol} ?$ (d) Using Figure 7.10 , determine which of the halogen elements has a first ionization energy closest to that of mercury.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

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