/*! 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} Q43E Suppose you discovered a diamond... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it?

Short Answer

Expert verified

The dissolution of silica or heating with \({\rm{CaO}}\)can help to chemically free adiamond encased in a silicate rock.

Step by step solution

01

Silicate rock

A rock that constitutes of silicate groups and is accountable for making 90% of the earth’s crust is called a silicate rock. Its application includes the manufacturing of ceramics, abrasives, and glass.

02

Process of chemically freeing the diamond without harming it

(a) A chemical way of removing the \({\rm{Si}}{{\rm{O}}_2}\) from the surface of the object would be by the dissolution of the silica. The only acid in which silica is soluble is hydrofluoric acid(HF), which would also react with diamond. The reaction is:

\({{\rm{H}}_4}{\rm{Si}}{{\rm{O}}_4} + 4HF \to SiF + 4{H_2}O.\)

(b) Another way would be by heating the object with \({\rm{CaO}}\). The reaction is:

\({\rm{CaO}}(s) + {\rm{Si}}{{\rm{O}}_2}(s) \to {\rm{CaSi}}{{\rm{O}}_3}(l).\)

The liquid product could be removed, and only the powder \({\rm{CaO}}\) would remain, which is soluble in water. This is possible because a diamond can stand very high temperatures without melting and reacting.

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 91Ó°ÊÓ!

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 a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding.

\((a)\;{{\rm{H}}_3}{\rm{BP}}{{\rm{H}}_3}\)

\((b)\;{\rm{BF}}_4^ - \)

\((c)\;{\rm{BB}}{{\rm{r}}_3}\)

\((d)\;{\rm{B}}{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}\)

\((e)\;{\rm{B}}{({\rm{OH}})_3}\)

What is the oxidation state of the halogen in each of the following?

\(\begin{array}{l}(a){{\rm{H}}_5}{\rm{I}}{{\rm{O}}_6}\\(b){\rm{IO}}_4^ - \\(c){\rm{Cl}}{{\rm{O}}_2}\\(d){\rm{IC}}{{\rm{l}}_3}\\(e){F_2}\end{array}\)

Basic solutions of \({\rm{N}}{{\rm{a}}_{\rm{4}}}{\rm{Xe}}{{\rm{0}}_{\rm{6}}}\) are powerful oxidants . What mass of \({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}{\rm{.6}}{{\rm{H}}_{\rm{2}}}{\rm{0}}\) reacts with 125.0 ml of 0.1717 M basic solution of \({\rm{N}}{{\rm{a}}_{\rm{4}}}{\rm{Xe}}{{\rm{0}}_{\rm{6}}}\) that contains an excess of Sodium Hydroxide if the products include Xe and the solution of Sodium Permanganate ?

A chemist dissolves a \({\bf{1}}.{\bf{497}} - {\bf{g}}\) sample of a type of metal (an alloy of \({\bf{Sn}},{\rm{ }}{\bf{Pb}},{\rm{ }}{\bf{Sb}},\)and\({\bf{Cu}}\)) in nitric acid, and metastannic acid,\({{\bf{H}}_2}{\bf{Sn}}{{\bf{O}}_3}\), is precipitated. She heats the precipitate to drive off the water, which leaves \({\bf{0}}.{\bf{4909}}\)g of tin (IV) oxide. What was the percentage of tin in the original sample?

Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding.

\(\begin{array}{l}(a){\rm{Xe}}{{\rm{F}}_2}\\(b){\rm{Xe}}{{\rm{F}}_4}\\(c){\rm{Xe}}{{\rm{O}}_3}\\(d){\rm{Xe}}{{\rm{O}}_4}\\(e){\rm{XeO}}{{\rm{F}}_4}\end{array}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.