TECTechnics Classroom TECTechnics Overview
Pj Problems - Overview
Celestial Stars
The Number Line
Geometries
7 Spaces Of Interest - Overview
Triadic Unit Mesh
Creation
The Atom
Survival
Energy
Light
Heat
Sound
Music
Language
Stories
Work
States Of Matter
Buoyancy
Nuclear Reactions
Molecular Shapes
Electron Configurations
Chemical Bonds
Energy Conversion
Chemical Reactions
Electromagnetism
Continuity
Growth
Human-cells
Proteins
Nucleic Acids
COHN - Natures Engineering Of The Human Body
The Human-Body Systems
Vision
Walking
Behaviors
Sensors Sensings
Beauty
Faith, Love, Charity
Photosynthesis
Weather
Systems
Algorithms
Tools
Networks
Search
Differential Calculus
Antiderivative
Integral Calculus
Economies
Inflation
Markets
Money Supply
Painting
A solution of 100 mL of 0.200 M KOH is mixed with a solution of 200 mL of 0.150 M NiSO4
(a) What precipitate forms?
(b) What is the limiting reactant?
(c) How many grams of the precipitate is formed?
(d) What is the concentration of each ion that remains in solution?
The strings:
S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).
The math:
Pj Problem of Interest is of type interaction (chemical).
(a) Chemical equation:
2KOH + NiSO4 ------> K2SO4 + Ni(OH)2
Precipitate is Ni(OH)2
(b) Mol of KOH = (0.100)(0.200) = 0.0200 mol
Mol of NiSO4 =(0.200)(0.150) = 0.03 mol
Mole ratio of KOH:NiSO4 in balanced equation is 2:1
Mole ratio of KOH:NiSO4 in solution is 1:1.5
So, KOH is the limiting reactant.
(c) 2 moles of KOH forms 1 mole of Ni(OH)2
So, 0.0200 mol of KOH forms 0.0200/2 = 0.01 mol of Ni(OH)2
Molar mass of Ni(OH)2 = 59 + 32 + 2 = 93 g
So, gram of precipitate = 93(0.01) = 0.93 g of Ni(OH)2.
(d) From calculations in (b):
Initial mol of KOH = 0.02 mol
So, inital mol of K
So, initial mol of OH- = 0.02
Initial mol of NiSO4 = 0.03 mol
So, initial mol of Ni2+ = 0.03
So, initial mol of SO42- = 0.03
Mole of Ni(OH)2 in product = 0.01
So mol of N2+ in product = 0.01
So, mol of 2OH- = 0.02
So, mol of Ni2+ that remain in solution = 0.03 - 0.01 = 0.02 mol
Combined volume of 0.100 L solution of KOH and 0.200 L solution of NiSO4 = 0.3 L
So, concentration of Ni2+ that remains in solution = 0.02/0.3 = 0.0667 M.
Initial mole of K+ remains in solution:
So, concentration of K+ that remains in solution = 0.02/0.3 = 0.0667 M.
Similarly, initial mol of SO42- remains in solution
So, concentraion of SO42- that remains in solution = 0.03/0.3 = 0.1 M.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
Single Variable Functions
Conics
Ordinary Differential Equations (ODEs)
Vector Spaces
Real Numbers
Separation Of Variables As Solution Method For Homogeneous Heat Flow Equation
Newton And Fourier Cooling Laws Applied To Heat Flow Boundary Conditions
Fourier Series
Derivation Of Heat Equation For A One-Dimensional Heat Flow
The Universe is composed of matter and radiant energy. Matter is any kind of mass-energy that moves with velocities less than the velocity of light. Radiant energy is any kind of mass-energy that moves with the velocity of light.
Periodic Table
Composition And Structure Of Matter
How Matter Gets Composed
How Matter Gets Composed (2)
Molecular Structure Of Matter
Molecular Shapes: Bond Length, Bond Angle
Molecular Shapes: Valence Shell Electron Pair Repulsion
Molecular Shapes: Orbital Hybridization
Molecular Shapes: Sigma Bonds Pi Bonds
Molecular Shapes: Non ABn Molecules
Molecular Orbital Theory
More Pj Problem Strings