/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 (a) Determine the number of calc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 77

(a) Determine the number of calcium ions in the chemical formula of the mineral hardystonite, \(\mathrm{Ca}_{x} \mathrm{Zn}\left(\mathrm{Si}_{2} \mathrm{O}_{7}\right)\). (b) Determine the number of hydroxide ions in the chemical formula of the mineral pyrophyllite, \(\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)_{2}(\mathrm{OH})_{x}\).

Short Answer

Expert verified
(a) There are 2 calcium ions in the chemical formula of the mineral hardystonite. (b) There is an error in the given formula for the mineral pyrophyllite, as the value of x should not be negative. Please verify the formula and resubmit the question.

Step by step solution

01

Identify the chemical formula of hardystonite

\(\mathrm{Ca}_{x} \mathrm{Zn}\left(\mathrm{Si}_{2} \mathrm{O}_{7}\right)\) is the chemical formula for the mineral hardystonite. To determine \(x (number\ of\ calcium\ ions)\), we need to know the charges of the ions in this formula.
02

Determine the charges of the ions in hardystonite

Calcium ion (Ca) has a charge of +2. Zinc ion (Zn) also has a charge of +2, and the silicate group \(\left(\mathrm{Si}_{2} \mathrm{O}_{7}\right)\) has a charge of -6.
03

Determine the balance of charges in hardystonite

Since hardystonite is a neutral compound, the sum of the charges from the ions must equal zero, we can now write an equation to determine the value of x: \(+2x\ +\ 2\ +\ (-6) = 0\) Step 2: Solve for x
04

Rearrange the equation and simplify

Rearrange the above equation to find the value of x: \(2x = 4\)
05

Find the value of x

Divide both sides of the equation by 2 to get the value of x: \(x = 2\) (a) So, there are 2 calcium ions in the chemical formula of the mineral hardystonite. Step 3: Determine the number of hydroxide ions in pyrophyllite
06

Identify the chemical formula of pyrophyllite

\(\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)_{2}(\mathrm{OH})_{x}\) is the chemical formula for the mineral pyrophyllite. To determine \(x\ (number\ of\ hydroxide\ ions)\), we need to know the charges of the ions in this formula.
07

Determine the charges of the ions in pyrophyllite

Aluminum ion (Al) has a charge of +3, and the silicate group \(\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)\) has a charge of -6. Hydroxide ion (OH) has a charge of -1.
08

Determine the balance of charges in pyrophyllite

As previously explained, the sum of the charges from the ions must equal zero for a neutral compound. We can now write an equation to determine the value of x: \(+6\ +\ (-12)\ +\ (-x) = 0\) Step 4: Solve for x
09

Rearrange the equation and simplify

Rearrange the above equation to find the value of x: \(-x = 6\)
10

Find the value of x

Multiply both sides of the equation by -1 to get the value of x: \(x = -6\) (b) So, there is an error in the given formula as the value of x should not be negative. Please verify the formula and resubmit the question.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Calcium Ions in Minerals
Minerals often contain calcium ions, which are essential components of their crystalline structures. Calcium ions have a +2 charge, which is essential for maintaining the overall charge balance in minerals. In the mineral hardystonite, with the formula \( \mathrm{Ca}_{x} \mathrm{Zn}(\mathrm{Si}_{2} \mathrm{O}_{7}) \), calcium ions help to neutralize the negative charge of the silicate group. By confirming the charges of involved ions and then setting up an equation to balance those charges, we understand that there are 2 calcium ions in hardystonite's formula.
Charge Balance in Compounds
Charge balance in compounds is a fundamental concept that ensures the overall electrical neutrality of a substance. All the positive and negative charges of the ions must add up to zero. In the exercise, the calculation relied on the fact that hardystonite as a neutral compound requires the sum of the charges from calcium, zinc, and the silicate group to be zero. This approach was also used to attempt to determine the number of hydroxide ions in the mineral pyrophyllite. Understanding the concept of charge balance helps in deducing the correct proportions of ions in the chemical formulae of minerals and compounds.
Hydroxide Ions in Minerals
Hydroxide ions ( \( \mathrm{OH}^- \)) play a crucial role in many minerals, especially in those classified as hydroxides. In the case of pyrophyllite, \( \mathrm{Al}_{2}(\mathrm{Si}_{2} \mathrm{O}_{5})_{2}(\mathrm{OH})_{x} \), hydroxide ions have a -1 charge which should balance out the positive charges of the aluminum ions. However, in the provided exercise, an error was detected since the number of hydroxide ions calculated yielded a negative charge. The presence and quantity of hydroxide ions can significantly affect the physical and chemical properties of minerals.

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

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group \(6 \mathrm{~A}\) element in each: (a) sulfur tetrachloride, (b) selenium trioxide, (c) sodium thiosulfate, (d) hydrogen sulfide, (e) sulfuric acid, (f) sulfur dioxide, (g) mercury telluride.

SOLUTION Analyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle. Plan We can best do this by writing a balanced nuclear reaction for the process. Solve The periodic table shows that radium has an atomic number of 88 . The complete chemical symbol for radium- 226 is therefore \({ }_{85}^{226} \mathrm{Ra}\). An alpha particle is a helium-4 nucleus, and so its symbol is \({ }_{2}^{4} \mathrm{He}\). The alpha particle is a product of the nuclear reaction, and so the equation is of the form $$ { }_{8}^{226} \mathrm{Ra} \longrightarrow{ }_{2}^{A} \mathrm{X}+{ }_{2}^{4} \mathrm{He} $$ where \(A\) is the mass number of the product nucleus and \(Z\) is its atomic number. Mass numbers and atomic numbers must balance, so $$ 226=A+4 $$ and $$ 88=Z+2 $$ Hence, $$ A=222 \text { and } Z=86 $$ Again, from the periodic table, the element with \(Z=86\) is radon (Rn). The product, therefore, is \({ }_{86}^{222} \mathrm{Rn}\), and the nuclear equation is $$ { }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} $$

The solubility of \(\mathrm{Cl}_{2}\) in \(100 \mathrm{~g}\) of water at \(\mathrm{STP}\) is \(310 \mathrm{~cm}^{3}\). Assume that this quantity of \(\mathrm{Cl}_{2}\) is dissolved and equilibrated as follows: $$ \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}(a q)+\mathrm{HClO}(a q)+\mathrm{H}^{+}(a q) $$ (a) If the equilibrium constant for this reaction is \(4.7 \times 10^{-4}\), calculate the equilibrium concentration of \(\mathrm{HClO}\) formed. (b) What is the \(\mathrm{pH}\) of the final solution?

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group \(6 \mathrm{~A}\) element in each: (a) selenous acid, (b) potassium hydrogen sulfite, (c) hydrogen telluride, (d) carbon disulfide, (e) calcium sulfate, (f) cadmium sulfide, (g) zinc telluride.

\text { Write the balanced nuclear equation for the process summarized as }{ }_{13}^{27} \mathrm{Al}(\mathrm{n}, \alpha)_{11}^{24} \mathrm{Na} \text {. }SOLUTION Analyze We must go from the condensed descriptive form of the reaction to the balanced nuclear equation. Plan We arrive at the balanced equation by writing \(\mathrm{n}\) and \(\alpha\), each with its associated subscripts and superscripts. Solve The \(\mathrm{n}\) is the abbreviation for a neutron \(\left({ }_{0}^{1} \mathrm{n}\right)\) and \(\alpha\) represents an alpha particle ( \(\left.{ }_{2}^{4} \mathrm{He}\right)\). The neutron is the bombarding particle, and the alpha particle is a product. Therefore, the nuclear equation is $$ { }_{13}^{27} \mathrm{Al}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{11}^{24} \mathrm{Na}+{ }_{2}^{4} \mathrm{He} \text { or } \quad{ }_{13}^{27} \mathrm{Al}+\mathrm{n} \longrightarrow{ }_{11}^{24} \mathrm{Na}+\alpha $$
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

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.