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Physiology Lab
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Muscarinic receptors would be found in which location(s) in the diagram
A. Heart Rate
B. TPR
C. EDV
D. SV
E. MAP
Increasing the viscosity of blood would directly affect:
A. TPR
B. CO
C. MAP
D. EDV
E. SV
The change in the radius of a vessel, raised to the fourth power can affect blood flow according to which principle Poiseuille’s Law. EDV changes directly impact what factor?
A. CO
B. TPR
C. MAP
D. SV
In the case of a hemorrhage, which of the following events might occur?
A. Vasoconstriction of vessels due to release Acetylcholine (Ach)
B. Decrease Heart Rate
C. Increase release Norepinephrine (NE) release
D. Elevated End Systolic Volume
Select all the structures in the flow diagram where Norepinephrine would act on alpha-1 receptor by clicking of structures in chart.
A. Venous return
B. Cardiac Output
C. TPR
D. SV
Muscarinic receptors are stimulated by:
A. Acetylcholine****
B. Norepinephrine
C. Nicotine
D. Gaba
E. Dopamine
Chemistry Lab
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Stoichiometry Exercise
In this laboratory exercise, you will need to understand how to use the masses of the elements/compounds to show the Law of Conservation of Mass in the decomposition reaction of a KCLO3.
The Stoichiometry laboratory exercise deals with the decomposition of potassium chlorate using manganese dioxide as a catalyst. The purpose of this activity is to conduct an experiment dealing with the decomposition of potassium chlorate in order to:
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To illustrate how the Law of Conservation of Mass is obeyed in a balanced chemical reaction.
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Identify the reactants and products in a balanced chemical reaction.
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Illustrate how to properly setup a chemical decomposition laboratory.
Develop and evaluate the quantitative nature of a chemical experiment using correct stoichiometrical relationships.
Step One: Use the Periodic Table provided to derive mass of each elements.
-Potassium (K) has a mass of 39
-Chlorine (Cl) has a mass of 35.5
-Oxygen (O) has a mass of 16.
Step Two: Add the atomic mass for each compound to demonstrate conservation of mass in the equation.
2KClO3
2K = 2 x 39g = 78g
2 Cl = 2 x 35.5g = 71g
6O = 6 x 16g = 96g
--------
245g
245g
2KCl + 3O2
2K = 2 x 39g = 78g
2 Cl = 2 x 35.5g = 71g
--------
149g
6O=6 x 16g = 96g
--------
+ 96g
= 245g
Step Three: Summary with illustration
In 2KClO₃ → 2KCl + 3O₂ we see that atoms and mass is conserved. But the large numbers of 2 : 2 : 3 represents the moles of each reactant, together they represent the molar ratio in the reaction but the moles are not conserved but are critical to stoichiometric calculation as you will see. Therefore the Law of conservation of Mass has been obeyed grams as a unit of mass.
Option to allow students to play with balancing equations
____Na + ____H20 → ____NaOH + ____H₂
____Zn + ____HF → ____ZnF2 + ____H₂
___ Mn(NO₂)₂ + ___ BeCl₂ ___ Be(NO₂)₂ + ___ MnCl₂
