The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined. Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons. Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound. The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it. The amount of substance containing 12 g of pure carbon is called a mole. One mole of atoms always contains 6 .022 × 10 23 atoms. Hence, ( 6 .022 × 10 23 atoms ) ( 12 u 1 atom ) = 12 g ⇒ 1 u = 1 6 .022 × 10 23 g To determine : The mass of phosphorous ( P ) in 5 .00 moles of P 4 O 6 .
The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined. Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons. Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound. The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it. The amount of substance containing 12 g of pure carbon is called a mole. One mole of atoms always contains 6 .022 × 10 23 atoms. Hence, ( 6 .022 × 10 23 atoms ) ( 12 u 1 atom ) = 12 g ⇒ 1 u = 1 6 .022 × 10 23 g To determine : The mass of phosphorous ( P ) in 5 .00 moles of P 4 O 6 .
Solution Summary: The author explains that the number of moles of a compound is given. The molar mass of any compound can be calculated by adding the atomic weight of individual atoms.
Study of body parts and their functions. In this combined field of study, anatomy refers to studying the body structure of organisms, whereas physiology refers to their function.
Chapter 5, Problem 58E
(a)
Interpretation Introduction
Interpretation: The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined.
Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.
Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.
The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.
The amount of substance containing
12g of pure carbon is called a mole. One mole of atoms always contains
6.022×1023 atoms.
Hence,
(6.022×1023atoms)(12u1atom)=12g⇒1u=16.022×1023g
To determine: The mass of phosphorous
(P) in
5.00 moles of
P4O6.
(b)
Interpretation Introduction
Interpretation: The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined.
Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.
Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.
The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.
The amount of substance containing
12g of pure carbon is called a mole. One mole of atoms always contains
6.022×1023 atoms.
Hence,
(6.022×1023atoms)(12u1atom)=12g⇒1u=16.022×1023g
(c)
Interpretation Introduction
Interpretation: The number of moles of compound is given. By using the number of moles, the mass of phosphorous in each compound given in exercise 52 is to be determined.
Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.
Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.
The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.
The amount of substance containing
12g of pure carbon is called a mole. One mole of atoms always contains
6.022×1023 atoms.
Hence,
(6.022×1023atoms)(12u1atom)=12g⇒1u=16.022×1023g
To determine: The mass of phosphorous
(P) in
5.00 moles of
Na2HPO4.
13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the
molecule depicted below.
Bond B
Bond A
Bond C
a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in
appropriate boxes.
Weakest
Bond
Strongest
Bond
b. (4pts) Consider the relative stability of all cleavage products that form when bonds A,
B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B,
and C are all carbon radicals.
i. Which ONE cleavage product is the most stable? A condensed or bond line
representation is fine.
ii. Which ONE cleavage product is the least stable? A condensed or bond line
representation is fine.
c. (5pts) Use principles discussed in lecture, supported by relevant structures, to
succinctly explain the why your part b (i) radical is more stable than your part b(ii)
radical. Written explanation can be no more than one-two succinct sentence(s)!
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Statifically more chances to abstract one of these 6H
11. (10pts total) Consider the radical chlorination of 1,3-diethylcyclohexane depicted below. 4
4th total
• 6H total
래
• 4H total
21 total
ZH
2H
Statistical
H < 3° C-H weakest
-
product
abstraction here
bund
leads to thermo favored
a) (6pts) How many unique mono-chlorinated products can be formed and what are the
structures for the thermodynamically and statistically favored products?
Product
6
Number of Unique
Mono-Chlorinated Products
Thermodynamically
Favored Product
Statistically
Favored Product
b) (4pts) Draw the arrow pushing mechanism for the FIRST propagation step (p-1) for the
formation of the thermodynamically favored product. Only draw the p-1 step. You do
not need to include lone pairs of electrons. No enthalpy calculation necessary
H
H-Cl
Waterfox
10. (5pts) Provide the complete arrow pushing mechanism for the chemical transformation →
depicted below
Use proper curved arrow notation that explicitly illustrates all bonds being broken, and
all bonds formed in the transformation.
Also, be sure to include all lone pairs and formal charges on all atoms involved in the
flow of electrons.
CH3O
II
HA
H
CH3O-H
H
①
Chapter 5 Solutions
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