<span>An element belonging to the halogen family would be expected to have a large ionization energy and a large electron affinity.
Flourine, Chlorine, Bromine, Iodine and astatine are the elements that belongs to the halogen family and mostly they have high values of ionization energy.
The amount of energy released when an electron is added to an atom or molecule to form a negative ion or anion is electron affinity.</span>Chlorine from this family has highest electron affinity.
Answer 2: 57.21235000000001
Answer 3: 44.01g
Answer 4: 24.645320000000005
Answer 5: 107.04578240000001
Good luck
Answer:
Exam 3 Material
Homework Page Without Visible Answers
This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
Explanation:
Answer:
180 amu
C₆H₁₂O₆
Explanation:
Step 1: Determine the molecular mass of the compound
The sample has a mass (m) of 3.06 g and it contains (n) 0.0170 moles. The molar mass M is:
M = m/n = 3.06/0.0170 mol = 180 g/mol
Then, the molecular mass is 180 amu.
Step 2: Determine the molar mass of the empirical formula.
M(CH₂O) = 1 × M(C) + 2 × M(H) + 1 × M(O)
M(CH₂O) = 1 × 12 g/mol + 2 × 1 g/mol + 1 × 16 g/mol = 30 g/mol
Step 3: Determine the molecular formula
First, we will determine "n" according to the following expression.
n = molar mass molecular formula / molar mass empirical formula
n = 180 g/mol / 30 g/mol = 6
The molecular formula is:
n × CH₂O = 6 × CH₂O = C₆H₁₂O₆
Answer:
Part A
The volume of the gaseous product is 
Part B
The volume of the the engine’s gaseous exhaust is 
Explanation:
Part A
From the question we are told that
The temperature is 
The pressure is 
The of 
The chemical equation for this combustion is
The number of moles of 
that reacted is mathematically represented as
The molar mass of is constant value which is
So 
The gaseous product in the reaction is 
and water vapour
Now from the reaction
2 moles of will react with 25 moles of 
to give (16 + 18) moles of and 
So
1 mole of will react with 12.5 moles of to give 17 moles of and
This implies that
0.8754 moles of will react with (12.5 * 0.8754 ) moles of to give (17 * 0.8754) of and
So the no of moles of gaseous product is
From the ideal gas law
making V the subject
Where R is the gas constant with a value 
Substituting values
Part B
From the reaction the number of moles of oxygen that reacted is
The volume is
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
Substituting values
