ELEM.PRIN.OF CHEMICAL PROC.-W/ACCESS
ELEM.PRIN.OF CHEMICAL PROC.-W/ACCESS
4th Edition
ISBN: 9781119330745
Author: FELDER
Publisher: WILEY
bartleby

Concept explainers

Question
Book Icon
Chapter 3, Problem 3.31P
Interpretation Introduction

(a)

Interpretation:

The volume of the API in mL to be sprayed for the patients with specified mass is to be calculated. Also, number of drops in every prescribed dose of the drug is to be calculated.

Concept introduction:

Mass of the drug (md) required for the given mass of patient (mP) and the prescribed dose (D) is:

md=mP×D   ........ (1)

Number of moles (n) is calculated by the formula:

n=mMW   ........ (2)

Here, m is the mass of the substance in 'g' and MW is the molecular mass of the substance in 'g/mol'.

Molarity (M) of any solution is calculated as:

M=nV   ........ (3)

Here, V is the volume of the solution in 'L'.

Number of drops (Nd) is calculated as:

Nd=VVd   ........ (4)

Here, V is the volume of the solution in 'mL' and Vd is the volume of one drop of the solution in 'mL'.

Volume of a sphere (Vs) is given by:

Vs=43πr3   ........ (5)

Here, r is the radius of the sphere in 'm'.

Interpretation Introduction

(b)

Interpretation:

The formula for the volume, Vdose(mL) in terms of the specified variables is to be derived. Also, the derived formula is to be checked for the values given in part (a).

Concept introduction:

Weight (W) in 'N' of an object is calculated as:

W=m×g   ........ (6)

Here, m is the mass of that object in 'kg' and g is the acceleration due to gravity, g=9.8 m/s2.

Mass of the drug (md) required for the given mass of patient (mP) and the prescribed dose (D) is:

md=mP×D   ........ (7)

Number of moles (n) is calculated by the formula:

n=mMW   ........ (8)

Here, m is the mass of the substance in 'g' and MW is the molecular mass of the substance in 'g/mol'.

Molarity (M) of any solution is calculated as:

M=nV   ........ (9)

Here, V is the volume of the solution in 'L'.

Interpretation Introduction

(c)

Interpretation:

For a sphere of radius ‘r’, its surface-to-volume ratio is to be calculated. Also, the surface area of the sphere for the specified volume and radius is to be determined.

Concept introduction:

Volume of a sphere (Vs) is given by:

Vs=43πr3   ........ (11)

Here, r is the radius of the sphere in 'm'.

Surface area of a sphere (SAs) is given by:

SAs=4πr2   ........ (12)

Here, r is the radius of the sphere in 'm'.

Blurred answer
Students have asked these similar questions
Question: McDaniel Shipyards wants to develop control charts to assess the quality of its steel plate. They... McDaniel Shipyards wants to develop control charts to assess the quality of its steel plate. They take ten sheets of 1" steel plate and compute the number of cosmetic flaws on each roll. Each sheet is 20' by 100'. Compute within 99.73% control limits. Based on the following data: a. Develop limits for the control chart b. Is the process in or out of control? c. Can you detect any outliers, if so which value(s)? Number of Sheet flaws 1 1 2 1 3 2 4 0 5 1 6 5 7 0 8 2 9 0 10 2
Question: McDaniel Shipyards wants to develop control charts to assess the quality of its steel plate. They take ten sheets of 1" steel plate and compute the number of cosmetic flaws on eac... McDaniel Shipyards wants to develop control charts to assess the quality of its steel plate. They take ten sheets of 1" steel plate and compute the number of cosmetic flaws on each roll. Each sheet is 20' by 100'. Based on the following data, develop limits for the control chart, plot the control chart, and determine whether the process is in control. Answer the following questions below. Number of flaws Sheet 1 1 2 1 = 3 2 4 0 5 1 6 5 7 0 8 2 9 10 0 2 PLEASE WRTIE NEATLY AND EXPLAIN! (: Thanks 1. Calculate the standard deviation of control chart. (a) the standard deviation = 1.0832 (b) the standard deviation = 1.1832 (c) the standard deviation = 1.4 (d) the standard deviation = 1.04 27. 2. Using +- 3 olimits, calculate the LCL and UCL for these data. 3.549; LCL = -3.549 (a) UCL (b) UCL 3.549;…
Derive an expression for incompressible flow in a horizontal pipe of constant diameter andwithout fittings or valves which shows that the pressure is a linear function of pipe length. Whatother assumptions are required for this result? Is this result valid for non-horizontal pipes? Howwill the presence of fittings, valves and other hardware affect this result?

Chapter 3 Solutions

ELEM.PRIN.OF CHEMICAL PROC.-W/ACCESS

Knowledge Booster
Background pattern image
Chemical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introduction to Chemical Engineering Thermodynami...
Chemical Engineering
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:McGraw-Hill Education
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemical Engineering
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Text book image
Elements of Chemical Reaction Engineering (5th Ed...
Chemical Engineering
ISBN:9780133887518
Author:H. Scott Fogler
Publisher:Prentice Hall
Text book image
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:9781119285915
Author:Seborg
Publisher:WILEY
Text book image
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:9781285061238
Author:Lokensgard, Erik
Publisher:Delmar Cengage Learning
Text book image
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:9780072848236
Author:Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:McGraw-Hill Companies, The