Molecular mass of Mg(OH)2
= Atomic mass of Mg + 2(Atomic mass of O) + 2(Atomic mass of H)
= 24 g/mol + 2(16 g/mol) + 2(1 g/mol )
= 58 g/mol
Molecular mass of Fe2O3
= 2(Atomic mass of Fe) + 3(Atomic mass of O)
= 2(56 g/mol) + 3(16 g/mol)
= 160 g/mol
Answer:
See below.
Explanation:
They both have 1 valency electron so will be metallic and in the same Group (Group 1) of the Periodic Table, so will have similar properties.
13.5g
Explanation:
Given parameters:
Mass of Na = 10g
Mass of O₂ = 10g
Unknown:
Mass of products formed = ?
Balanced equation = ?
Solution:
The balanced chemical equation is shown below:
4Na + O₂ ⇒ 2Na₂O
In any reaction, the specie in short supply determines the extent of the reaction.
This reaction is not an exclusion. We need to first determine the specie in short supply and use it to estimate the amount of product since we have a 100% yield which signifies that all was used up.
let us convert to moles;
Number of moles of Na = 
= 0.435mole
Number of moles of O₂ = 
= 0.313mole
From the given equation;
4 moles of Na requires 1 mole of O₂;
0.435 moles of Na will require 
= 0.11 moles
But the given amount O₂ is 0.313, this is an excess of 0.313 - 0.11 = 0.203moles
We see that Na is the limiting reagent;
4 moles of Na gives 2 mole of Na₂O
0.435 moles of Na will give 
= 0.22 moles
Mass of Na₂O = number of moles x molar mass = 62 x 0.22 = 13.5g
learn more:
Number of moles brainly.com/question/1841136
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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:
The predominant intermolecular force in the liquid state of each of these compounds:
ammonia (NH3)
methane (CH4)
and nitrogen trifluoride (NF3)
Explanation:
The types of intermolecular forces:
1.Hydrogen bonding: It is a weak electrostatic force of attraction that exists between the hydrogen atom and a highly electronegative atom like N,O,F.
2.Dipole-dipole interactions: They exist between the oppositely charged dipoles in a polar covalent molecule.
3. London dispersion forces exist between all the atoms and molecules.
NH3 ammonia consists of intermolecular H-bonding.
Methane has London dispersion forces.
Because both carbon and hydrogen has almost similar electronegativity values.
NF3 has dipole-dipole interactions due to the electronegativity variations between nitrogen and fluorine.
