Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 68

Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water, (c) chromium(III) oxide with dilute hydrochloric acid, (d) selenium dioxide with aqueous potassium hydroxide.

Short Answer

Expert verified
The balanced equations for the given reactions are: a) \(K_2O + H_2O \to 2KOH\) b) \(P_4O_6 + 6H_2O \to 4H_3PO_3\) c) \(Cr_2O_3 + 6HCl \to 2CrCl_3 + 3H_2O\) d) \(SeO_2 + 2KOH \to K_2SeO_3 + H_2O\)

Step by step solution

01

a) Potassium oxide with water

Potassium oxide (K2O) reacts with water (H2O) to produce potassium hydroxide (KOH). The balanced equation for this reaction can be found by ensuring that the number of atoms of each element is equal on both sides of the equation: \[ K_2O + H_2O \to 2KOH \]
02

b) Diphosphorus trioxide with water

Diphosphorus trioxide (P4O6) reacts with water (H2O) to form phosphorous acid (H3PO3). To balance the equation, ensure the number of atoms of each element is equal on both sides: \[ P_4O_6 + 6H_2O \to 4H_3PO_3 \]
03

c) Chromium(III) oxide with dilute hydrochloric acid

Chromium(III) oxide (Cr2O3) reacts with dilute hydrochloric acid (HCl) to form chromium(III) chloride (CrCl3) and water (H2O). Balance the equation to ensure the same number of atoms of each element on both sides: \[ Cr_2O_3 + 6HCl \to 2CrCl_3 + 3H_2O \]
04

d) Selenium dioxide with aqueous potassium hydroxide

Selenium dioxide (SeO2) reacts with aqueous potassium hydroxide (KOH) to form potassium selenite (K2SeO3) and water (H2O). Ensure the number of atoms of each element is equal on both sides to balance the equation: \[ SeO_2 + 2KOH \to K_2SeO_3 + H_2O \]

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

Moseley established the concept of atomic number by studying X-rays emitted by the elements. The X-rays emitted by some of the elements have the following wavelengths: $$ \begin{array}{ll} \hline \text { Element } & \text { Wavelength }(\AA) \\\ \hline \mathrm{Ne} & 14.610 \\ \mathrm{Ca} & 3.358 \\ \mathrm{Zn} & 1.435 \\\ \mathrm{Zr} & 0.786 \\ \mathrm{Sn} & 0.491 \\ \hline \end{array} $$

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Cesium is added to water. (b) Strontium is added to water. (c) Sodium reacts with oxygen. (d) Calcium reacts with iodine.

Discussing this chapter, a classmate says, "An element that commonly forms a cation is a metal." Do you agree or disagree? Explain your answer.

Write the electron configurations for the following ions: (a) \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Hg}^{2+},(\mathbf{c}) \mathrm{Mn}^{2+},(\mathrm{d}) \mathrm{Pt}^{2+},(\mathrm{e}) \mathrm{P}^{3-}\).

Until the early 1960 s the group 8 A elements were called the inert gases; before that they were called the rare gases. The term rare gases was dropped after it was discovered that argon accounts for roughly \(1 \%\) of Earth's atmosphere. (a) Why was the term inert gases dropped? (b) What discovery triggered this change in name? (c) What name is applied to the group now?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

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