Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 12, Problem 12.128P

a.

Interpretation Introduction

Interpretation:

The number of wafers that can be made from Si ingot is to be calculated.

Concept introduction:

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is kg/m3. The formula to calculate density is,

  Density=MassVolume        (1)

The formula to calculate the volume of the cylinder is as follows:

  V=πr2h        (2)

Here,

r is the radius of the cylinder.

h is the height of the cylinder.

a.

Expert Solution
Check Mark

Answer to Problem 12.128P

The number of wafers that can be made from Si ingot is 1.11×103.

Explanation of Solution

Rearrange the equation (1) to calculate the volume of Si ingot.

  VolumeofSi ingot=(MassDensity)        (3)

Substitute 2.34g/cm3 for density and 4.00kg for mass in the equation (3) to calculate the volume of Si ingot.

  Volumeof Siingot=(4.00kg2.34g/cm3)(1000g1kg)=1709.402cm3

The diameter of Si ingot is 5.20in. Therefore the radius of Si ingot is calculated as follows:

Radius ofSi ingot=(5.20in2)=2.60in(2.54cm1in)=6.604 cm

Rearrange the equation (2) to calculate the height of Si ingot.

  h=Vπr2        (4)

Substitute 1709.402cm3 for V, 3.14159 for π and 6.604 cm for r in the equation (4) to calculate the height of Si ingot.

  h=1709.402cm3(3.14159)(6.604 cm)2=12.476 cm

The thickness of the wafer is 1.12×104m. Therefore the number of the wafer is calculated as follows:

  Number of wafer=(12.476 cm1.12×104m)(1m100cm)=1113.94=1.11×103.

Conclusion

1.11×103 Si ingot can be made from 4.00kg cylindrical ingot that is 5.20in in diameter.

(b)

Interpretation Introduction

Interpretation:

The mass of a wafer is to be calculated.

Concept introduction:

The formula to calculate the mass of the cylinder is as follows:

  M=πr2hd        (5)

Here,

M is the mass of the cylinder.

r is the radius of the cylinder.

h is the height of the cylinder.

d is the density of the cylinder.

(b)

Expert Solution
Check Mark

Answer to Problem 12.128P

The mass of a wafer is 3.59g.

Explanation of Solution

Substitute 1.12×104m for h, 3.14159 for π and 6.604 cm for r, 2.34 g/cm3 for d in the equation (5) to calculate the mass of wafer.

  M=(3.14159)(6.604 cm)2(1.12×104m)(100cm1m)(2.34 g/cm3)=3.5909 g3.59g.

Conclusion

The shape of the wafer is cylindrical. The mass of a wafer is calculated as 3.59g.

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the removal of the oxide layer on the wafer is to be written.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

Displacement reactions are those in which substances on both sides of the equation remain the same but the atoms exchange places in order to form the product. Displacement reactions can be further classified as a double displacement reaction and a single displacement reaction. In double displacement reaction, both the reactants are the compounds and in single displacement reaction, one of the reactants is an element. The general representation of a double displacement reaction is:

  AB+CDAD+CB

(c)

Expert Solution
Check Mark

Answer to Problem 12.128P

The balanced equation for the removal of the oxide layer on the wafer is as follows:

  SiO2(s)+4HF(g)SiF4(g)+2H2O(g)

Explanation of Solution

Silicon combines with oxygen to form silicon dioxide. The coating of silicon dioxide interferes with the function of the wafer and is removed by gaseous hydrogen fluoride.

Sulfur displaces the hydrogen from hydrogen fluoride and leads to the formation of silicon tetrafluoride and water. The chemical equation for the reaction is as follows:

  SiO2(s)+4HF(g)SiF4(g)+2H2O(g)

Conclusion

Displacement reactions are those in which substances on both sides of the equation remain the same but the atoms exchange places in order to form the product.

(d)

Interpretation Introduction

Interpretation:

The moles of HF required per wafer is to be calculated.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  A+2B3C

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

(d)

Expert Solution
Check Mark

Answer to Problem 12.128P

The moles of HF required per wafer is 3.84×103mol.

Explanation of Solution

The formula to calculate the moles of HF is as follows:

  Moles ofHF=mass ofSimolar mass of Si(0.750%100%)(4molHF1molSi)        (6)

Substitute 3.5909g for the mass of Si, 28.09g/mol for molar mass of Si in the equation (6).

  Moles ofHF=3.5909g28.09g/mol(0.750%100%)(4molHF1molSi)=3.83506×103mol3.84×103mol.

Conclusion

The number of moles of water required by one mole of wafer is 3.84×103mol.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 12 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 12.6 - For each of the following crystalline solids,...Ch. 12.6 - Prob. 12.6BFPCh. 12.6 - Prob. 12.7AFPCh. 12.6 - Iron crystallizes in a body-centered cubic...Ch. 12.6 - Prob. 12.8AFPCh. 12.6 - Prob. 12.8BFPCh. 12.6 - Prob. B12.1PCh. 12.6 - Prob. B12.2PCh. 12 - Prob. 12.1PCh. 12 - Prob. 12.2PCh. 12 - Prob. 12.3PCh. 12 - Prob. 12.4PCh. 12 - Prob. 12.5PCh. 12 - Prob. 12.6PCh. 12 - Prob. 12.7PCh. 12 - Name the phase change in each of these events: (a)...Ch. 12 - Prob. 12.9PCh. 12 - Many heat-sensitive and oxygen-sensitive solids,...Ch. 12 - Prob. 12.11PCh. 12 - Prob. 12.12PCh. 12 - Prob. 12.13PCh. 12 - Prob. 12.14PCh. 12 - Prob. 12.15PCh. 12 - Prob. 12.16PCh. 12 - Prob. 12.17PCh. 12 - Prob. 12.18PCh. 12 - From the data below, calculate the total heat (in...Ch. 12 - Prob. 12.20PCh. 12 - Prob. 12.21PCh. 12 - Prob. 12.22PCh. 12 - Prob. 12.23PCh. 12 - Prob. 12.24PCh. 12 - Prob. 12.25PCh. 12 - Prob. 12.26PCh. 12 - Prob. 12.27PCh. 12 - Prob. 12.28PCh. 12 - Prob. 12.29PCh. 12 - Prob. 12.30PCh. 12 - Use Figure 12.10 to answer the following: Carbon...Ch. 12 - Prob. 12.32PCh. 12 - Prob. 12.33PCh. 12 - Prob. 12.34PCh. 12 - Prob. 12.35PCh. 12 - Prob. 12.36PCh. 12 - Distinguish between polarizability and polarity....Ch. 12 - Prob. 12.38PCh. 12 - Prob. 12.39PCh. 12 - Prob. 12.40PCh. 12 - Prob. 12.41PCh. 12 - Prob. 12.42PCh. 12 - Prob. 12.43PCh. 12 - Prob. 12.44PCh. 12 - Prob. 12.45PCh. 12 - Prob. 12.46PCh. 12 - Prob. 12.47PCh. 12 - Prob. 12.48PCh. 12 - Prob. 12.49PCh. 12 - Which liquid in each pair has the lower vapor...Ch. 12 - Which substance has the lower boiling point?...Ch. 12 - Which substance has the higher boiling point?...Ch. 12 - Prob. 12.53PCh. 12 - Prob. 12.54PCh. 12 - Prob. 12.55PCh. 12 - Prob. 12.56PCh. 12 - Why does the antifreeze ingredient ethylene glycol...Ch. 12 - Prob. 12.58PCh. 12 - Prob. 12.59PCh. 12 - Why does an aqueous solution of ethanol (CH3CH2OH)...Ch. 12 - Prob. 12.61PCh. 12 - Prob. 12.62PCh. 12 - Prob. 12.63PCh. 12 - Prob. 12.64PCh. 12 - Prob. 12.65PCh. 12 - Prob. 12.66PCh. 12 - Prob. 12.67PCh. 12 - Prob. 12.68PCh. 12 - Prob. 12.69PCh. 12 - Prob. 12.70PCh. 12 - Prob. 12.71PCh. 12 - Prob. 12.72PCh. 12 - Prob. 12.73PCh. 12 - Prob. 12.74PCh. 12 - Prob. 12.75PCh. 12 - Prob. 12.76PCh. 12 - Prob. 12.77PCh. 12 - Prob. 12.78PCh. 12 - Prob. 12.79PCh. 12 - Prob. 12.80PCh. 12 - Prob. 12.81PCh. 12 - Prob. 12.82PCh. 12 - Prob. 12.83PCh. 12 - Prob. 12.84PCh. 12 - Besides the type of unit cell, what information is...Ch. 12 - What type of unit cell does each metal use in its...Ch. 12 - What is the number of atoms per unit cell for each...Ch. 12 - Calcium crystallizes in a cubic closest packed...Ch. 12 - Chromium adopts the body-centered cubic unit cell...Ch. 12 - Prob. 12.90PCh. 12 - Prob. 12.91PCh. 12 - Prob. 12.92PCh. 12 - Prob. 12.93PCh. 12 - Prob. 12.94PCh. 12 - Prob. 12.95PCh. 12 - Prob. 12.96PCh. 12 - Prob. 12.97PCh. 12 - Prob. 12.98PCh. 12 - Prob. 12.99PCh. 12 - Prob. 12.100PCh. 12 - Prob. 12.101PCh. 12 - Prob. 12.102PCh. 12 - Prob. 12.103PCh. 12 - Polonium, the Period 6 member of Group 6A(16), is...Ch. 12 - Prob. 12.105PCh. 12 - Prob. 12.106PCh. 12 - Prob. 12.107PCh. 12 - Prob. 12.108PCh. 12 - Prob. 12.109PCh. 12 - Prob. 12.110PCh. 12 - Prob. 12.111PCh. 12 - Prob. 12.112PCh. 12 - Prob. 12.113PCh. 12 - Prob. 12.114PCh. 12 - Prob. 12.115PCh. 12 - Prob. 12.116PCh. 12 - Prob. 12.117PCh. 12 - Prob. 12.118PCh. 12 - Prob. 12.119PCh. 12 - Prob. 12.120PCh. 12 - Prob. 12.121PCh. 12 - Prob. 12.122PCh. 12 - Prob. 12.123PCh. 12 - Prob. 12.124PCh. 12 - Prob. 12.125PCh. 12 - Prob. 12.126PCh. 12 - Bismuth is used to calibrate instruments employed...Ch. 12 - Prob. 12.128PCh. 12 - Prob. 12.129PCh. 12 - Prob. 12.130PCh. 12 - Prob. 12.131PCh. 12 - Prob. 12.132PCh. 12 - Prob. 12.133PCh. 12 - Prob. 12.134PCh. 12 - Prob. 12.135PCh. 12 - Prob. 12.136PCh. 12 - Prob. 12.137PCh. 12 - Prob. 12.138PCh. 12 - Prob. 12.139PCh. 12 - Prob. 12.140PCh. 12 - Prob. 12.141PCh. 12 - Prob. 12.142PCh. 12 - Prob. 12.143PCh. 12 - Prob. 12.144PCh. 12 - Prob. 12.145PCh. 12 - The crystal structure of sodium is based on the...Ch. 12 - Prob. 12.147PCh. 12 - One way of purifying gaseous H2 is to pass it...
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Types of Matter: Elements, Compounds and Mixtures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=dggHWvFJ8Xs;License: Standard YouTube License, CC-BY