EBK ELEMENTARY PRINCIPLES OF CHEMICAL P
EBK ELEMENTARY PRINCIPLES OF CHEMICAL P
4th Edition
ISBN: 9781119192107
Author: BULLARD
Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT
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Chapter 3, Problem 3.20P
Interpretation Introduction

(a)

Interpretation:

How many of kg of C8 H10 in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

Interpretation Introduction

(b)

Interpretation:

How many of mol of C8 H10 in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

Units can be converted as,

1 kmol = 1000 mol

Interpretation Introduction

(c)

Interpretation:

How many of lb-mol of C8 H10 in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

Units can be converted as,

1 lb-mol = 453.6 mol

Interpretation Introduction

(d)

Interpretation:

How many of mol (g-atom) of C in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

Interpretation Introduction

(e)

Interpretation:

How many of mol (g-atom) of H in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

Interpretation Introduction

(f)

Interpretation:

How many of g of C in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

Interpretation Introduction

(g)

Interpretation:

How many of g of H in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

Interpretation Introduction

(h)

Interpretation:

How many of molecules in 15 kmol of C8 H10.

Concept introduction:

The mole is defined as the amount of substance containing the same number of chemical units. This can be written as,

Mole = Mass of the substance / Molecular weight of the substance

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Students have asked these similar questions
3. (a) If 10.0 mL of H2SO4 (sp. Gr. 1.50, containing 48.7% of combined SO3 by weight) is diluted to 400 mL, what is the normality of the solution as an acid? (b) What volume of 6.00 M H2SO4 should be added to this in order to make the resulting mixture 1.00 N as an acid? FW's : H2SO4 = 98; SO3 = 80
If 10.0 mL of H2SO4 (sp. Gr. 1.50, containing 48.7% of combined SO3 by weight) is diluted to 400 mL, what is the normality of the solution as an acid? (b) What volume of 6.00 M H2SO4 should be added to this in order to make the resulting mixture 1.00N as an acid? FW's : H2SC4 = 98; SO3 = 80
II. Determining the Volume of 0, Produced from the H20, Solution code number of H2O, solution Equate 1.02 density of H,O2 solution, g mL¯1 3 mass percent H,O2 in H2O2 solution, % determination 1 2 3 PV(H2O)=20. 070 5 5 5 volume of H2O2 solution used, mL 22.22 22.22 22.2 water temperature, °C 29.55 29.55 29.55 barometric pressure, in. Hg barometric pressure, torr BP = (P in. Hg) (25.4 torr/1 in.Hg) pressure of collected O2 at water temperature, torr P(02) = BP-P(H2O VARRE-temperature, K 49.3 50.2 50.1 second buret reading, mL first buret reading, mL volume of O, collected at laboratory T and P, mL P(H2O vap) = 15.5

Chapter 3 Solutions

EBK ELEMENTARY PRINCIPLES OF CHEMICAL P

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